麻豆精品视频在线观看视频,黄色小说综合网站,国内外成人免费激情在线视频

poj1207(水题不水�?--RMQ�Q�线�D�|��(w��i))

wangs — Sat, 28 Apr 2012 08:15:00 GMT

http://poj.org/problem?id=1207

		3. 		 if n = 1 then STOP 
		4. 		 		 if n is odd then   n <-- 3n+1 
		5. 		 		 else   n <-- n/2
问题很明白，之前在hdu上也做了�Q�经验还是没有哇。WA了好多好多次�?br />1w的数据很弱，直接暴力当然能过�?br />不过100W��׃��行了。记忆化搜烦(ch��)DP是个好办�?br />我最之前的做法就是这个，效果�q�算不错�?br />
求出所有的值后���是区间求最大��g���Q�有RMQ���法�Q�线�D�|��(w��i)�{�都行。不�q�，我都不会(x��)呀�Q�！�Q�！�Q�！�Q�！
朗讯的时候就做了个区间最值的问题�Q�当时朴素算法一直WA�Q�jh辉神 qsort()一下，从前往后给AC了。Ｙ�Q�Ｙ�Q?br />
最后还是就是细节问题了�Q�像输出ｓ，ｔ。有可能ｓ＞ｔ，输出却是要按照输入顺序输出！晕，���p����P�D�Q�了一天�?br />�ȝ���Q?br />      �l�节问题要注意！多想惛_��他算法，水题不水呀�?br />

#include
#include
#include
long long a[10005];
int GetC()
{
    int i,t;
    for (i=1;i<=10000 ;i++ )
    {
        t=i;a[i]=1;
        while (t!=1)
        {
            a[i]++;
            if (t&1)
                t=3*t+1;
            else
                t/=2;
        }
    }
}
int main()
{
    int s,t,i;
    long long mm;
    GetC();
    while (scanf("%d%d",&s,&t)==2)
    {
        printf("%d %d ",s,t);
        if (s>t)
            i=s,s=t,t=i;
        mm=a[s];
        for (i=s;i<=t ;i++ )
            mm=mm>a[i]?mm:a[i];
        printf("%I64d\n",mm);
    }
    return 0;
}
#include
#include
#include
long long a[10005];
int GetC()
{
    int i,t;
    for (i=1;i<=10000 ;i++ )
    {
        t=i;a[i]=1;
        while (t!=1)
        {
            a[i]++;
            if (t&1)
                t=3*t+1;
            else
                t/=2;
        }
    }
}
int main()
{
    int s,t,i;
    long long mm;
    GetC();
    while (scanf("%d%d",&s,&t)==2)
    {
        printf("%d %d ",s,t); //�q�个WA了一��?br />        if (s>t)                    //�q�个WAl了好多次�Q�！
            i=s,s=t,t=i;
        mm=a[s];
        for (i=s;i<=t ;i++ )
            mm=mm>a[i]?mm:a[i];
        printf("%I64d\n",mm);
    }
    return 0;
}

额，代码�q�是没有自己风格啊。这些个函数名还是那么难取呢�?br />

wangs 2012-04-28 16:15 发表评论

poj 2635(高精度求�?同余模定�?

wangs — Mon, 16 Apr 2012 13:29:00 GMT

http://poj.org/problem?id=2635

题目大意�Q�给定一个大整数�Q�这个数是两个素数的乘积�Q�然后给定一个数L�Q�如果这两个素数中有一个比L���，���p��出BAD�Q�不然输出GOOD

高精度求�?同余模定理：(x��)

高精度求模，一边加一�Ҏ(gu��)��Q�不�q�如果是100或�?000�q�制�q�些�Q�要注意啦，�?000�q�制 12345 应该�?2 345�q�样�Q�不要弄�E?23 45。我��是�q�样错的哦！�Q�！
同余模定理：(x��)http://hi.baidu.com/koomo007/blog/item/110cd6f58bc91964ddc47424.html

额，W(xu��)A了很多次哈，先是没有惛_��?0�q�制做要TLE�Q�后来用1000有搞错了�Q�！�Q?br />
�ȝ��Q�各�U�方法算法要吃透才行啊。不能只领略大概。计��好旉��I�间复杂度！
WA 3�ơ，1�ơAC 954MS

#include<stdio.h>
#include<string.h>
#include<math.h>
char ch[155];
int prime[500005],b[1010000],a[155];
int n,tot,m;
int get_mod(int p)
{
    int i,ans;
    ans=0;
    for (i=1; i<=m ; i++ )
        ans=(ans*1000+a[i])%p;
    return ans;
}
int main()
{
    int i,j,k,t,len;
    memset(b,0,sizeof(b));
    i=2;
    tot=0;
    while (1)
    {
        while (b[i])    i++;
        prime[++tot]=i;
        j=i;
        if (i>1000000)
            break;
        while (j<=1010000)
        {
            b[j]=1;
            j+=i;
        }
    }

    while (scanf("%s%d",&ch,&n)==2&&n)
    {
        m=strlen(ch);
        i=0;
        k=0;
        t=m%3;
        if (t>0)
        {
            a[++k]=0;
            for (i=0; i<t ; i++ )
                a[k]=a[k]*10+ch[i]-48;
        }
        while (i<m)
        {
            t=3<m-i?3:m-i;
            a[++k]=0;
            for (j=i; j<i+t ; j++ )
            {
                a[k]=a[k]*10+ch[j]-48;
            }
            i+=t;
        }
        m=k;

        i=1;
        while (i<=tot&&prime[i]<n&&get_mod(prime[i]))
            i++;

        if (prime[i]<n)
            printf("BAD %d\n",prime[i]);
        else
            printf("GOOD\n");
    }
    return 0;
}

wangs 2012-04-16 21:29 发表评论

中国��h��大学�W�一�?华�ؓ(f��)�?�E�序设计竞赛(高��l?

wangs — Mon, 16 Apr 2012 10:23:00 GMT

本来是和毛哥�l�好了队��d��题的�Q�哎�Q�毛哥有译֕��Q�耽误不得�?br />压力山大了ing…………

热��n赛说是去�q�朗讯的高��l�的题目�Q�呵呵，虽然�q�有一题没有做出来�Q�不�q�，�q�是��L��交了yes了四个。。。中途网�l�出了点问题�Q�感觉华为的�q�是没有�l�验�Q�呵呵，�W�一�ơ嘛�?br />居然通知说高�U�组题目全部是英语，晕了�Q�得找个��译啊。哈哈，王惟伊同学当然是首选啦�Q�不想hxj她去被打��M��^.^

惟伊同学�q�真是给力，哈哈�Q�上�ơ联赛就��L��虐场�Q�这�ơ一起心态很好很好^

额，比赛�Q�心态很重要哦�?br />
5个题目，之前我就�l�安排了先看题目短的题目。第一个题目是巨长的描�q�啊�Q�果断看其他的�?br />我拣了个数据比较��单的C题看了看�Q�哎�Q�贪心，so easy。还用上了我刚刚写的qsort.
�W�一个提交就yes了，呵呵�Q�第一个哈。应该不��过15min吧，挺爽的�?br />
惟伊同学�l�了我B(t��i)题，描述完了�Q�呵呵，dp嘛，方程一下子��出来了。写完样例就�q�了�Q�果断提交。返回了个NO�Q�想到肯定是��界溢出了，int不够。换long long�Q�结果编译错误，哎，服务器不行啊。只有换double了，呵呵�Q�yes了。不�q�开哥说double后面��׃��_��了，�q�挺不放心的�Q�看来是惛_��余了。得好好研究c才行�?

兴奋兴奋�Q�两个题�Q�不怕不被虐了。现在大概过�?0min了吧�Q?br />
然后是D题，惟伊同学说也是个dp�Q�哈哈哈�Q?1串计数问题，当然dp啦�?br />不过�q�里�U�结了挺久，原来脑中一开始的那个方程是有问题的，�U�结的挺久。以后做dp�Q�还是得先写好方�E�和状态表�C�，不要先写代码�?br />
样例�q�了�Q�嘿嘿，我聪明了下，��(g��)查了看看有没有越界，呵呵�Q�果然越界溢��Z��。果断地换了double�?br />一下yes!!哈哈�Q?个题目，已经�q�上辉哥啦，80min�q�去�?br />惟伊同学问了问开哥，呵呵�Q�only 1 yes�Q�额�Q�开哥没发挥好�?br />
后面��?zh��n)�惨了�Q�两个半多小时做最后一题，bfs�Q�那个状态和转移啊，一开始就没有设计好，考虑全面�Q�结果边写边设计�Q�自��p��力还是不够，��费了不��时��_(d��)��后来又出问题在方向上�Q�哎�Q��?zh��n)�惨了�Q�bfs�Q�dfs都应该想��x��向啊�Q�字典序�q�些�Q�。等搞定�q�些�Q�就没有剩下旉��啦，最后去看发现自己居焉��列开��了�Q�之前算�?�Q�不对，应该�?�Q�啊�Q�坑爹坑爹呢。。。。。�?br />
�W�一题就不说了，惟伊同学都没有搞清楚题意�Q�主要是输出那个表示没有搞明白，呵呵呵，不过��x��其实也是个状态压�~�的��单题目�?br />
5个题目，AC3个，遗憾一个。这�ơ华为杯��是个热�w�吧�Q�省赛才是重点！�Q�！

A�Q�状态压�~?br />B:dp
C:贪心
D:dp
E�Q�bfs�Q�字典序�Q?br />

wangs 2012-04-16 18:23 发表评论

hdu(位运��?)

wangs — Fri, 13 Apr 2012 16:43:00 GMT

�Q?413�Q�群赛里的最后一题，有点隑ֺ��?br />�W�一��D��得挺��单的�Q?000k的数�Q�想先给它排序然后找出来。哎�Q�这不是sb的做法嘛�Q�那么朴素的��法�Q�纯�_�的找虐�Q�！jh别写了，1M的空间写个毛啊�?br />��x��Q�愣了。hash也是不行�Q�没有办法，jh说肯定是啥高�U�数据结构来做了�Q�嗯�Q�我们就是很多高�U�数据结构不�?x��)，哎，伤心^�Q�。其实现在想惻I��1M也就250K个int敎ͼ�极端情况�?00K是完全没有办法处理的�Q�看来高�U�数据结构也不行了（如果再大点空��_(d��)��估计二叉查找�?w��i)和zikai学长说的set是可以实现的哈）(j��)

额，惛_��?*n+2,�q�个数模3*n�Q�就是剩下的那两个数�?*n了，直接�q�样也是没有办法来处理的。该怎么改进好呢�Q�得再研�I�研�I�^……^

异或�q�算�Q�呵呵，辉哥惛_��q�个可以。嗯�Q�也是，演算了一下，��L��里学�q�的几个�q�算可以实现把a@a@a�q�种�l�处理掉成单位元�?#8230;…�q�里接着��x��才行�?br />
原来位运��实现可以分解到2�q�制来做�Q�模拟。哈哈，真是好东�ѝ�?br />a[],b[],c[][]�Q�这里a[i]表示�q�些数分解到�W�i位上的篏加，�?之后��是那两个数在这个位上的��g��。c[i][j]表示i位和j位是否在某个��C��?br />�q�样之后如果a[i]==2那这两个数都在这个位上有分解�Q�各自篏加上去，
如果a[i]==1��得讨论了，如果记当前分解的数在一个有分解的位�|�是flag�Q�则如果c[flag][i]==1那么可以知道i位也是这个数的分解（ps,�q�里c[flag][i]不会(x��)�?的）(j��)
额，最�q�在看群��Z��么的�Q�想��C��一映射�Q�双��）(j��)�Q�这些个好理��是挺有用的哈。。�?br />
jh是用了s和s^2分别地映��过去，�q�样��出来x+y=t1,x^2+y^2=t2,�q�个方程好解的�?br />
�ȝ��Q?br /> 以后做题要看看数据范��_(d��)��旉��I�间。先设计好算法，先估计好复杂度才行！�Q?br /> hdu上一�ơAC�Q�运行了1000+Ms.不知道那�?Ms的是怎么出来�l�果的，求解�Q?br />��赛AC代码�Q�（�W�一�U�方法）(j��)

hdu AC代码�Q�（�W�二�U�方法）(j��)

#include<stdio.h>
#include<string.h>
#include<math.h>
long long x,y;
int calc(int a[],long long s)
{
    int i;
    i=0;
    while (s)
    {
        i++;
        if (s%2==1)
        {
            a[i]++;
        }
        s=s/2;
    }
    return 0;
}
int main()
{
    int t,i,j,n;
    int a[70],b[70];
    long long x,y,s;
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d",&n);
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for (i=1; i<=n ; i++ )
        {
            scanf("%I64d",&s);
            calc(a,s);
            calc(b,s*s);
        }
        for (i=1; i<=65 ; i++ )
            a[i]%=3;
        x=0;

        for (i=65; i>=1 ; i-- )
        {
            x*=2;
            x+=a[i];
        }
        for (i=1; i<=65 ; i++ )
            b[i]%=3;
        y=0;
        for (i=65; i>=1 ; i-- )
        {
            y*=2;
            y+=b[i];
        }

        printf("%.0lf %.0lf\n",(double)(x-sqrt((double)2*y-x*x))/2.0,(double)(sqrt((double)2*y-x*x)+x)/2.0);
    }
    return 0;
}

额，C很弱�Q�得好好看看Brian W.Kernighan和Dennis M. Ritchie的《C Programming Language》。再多了解了解编译器和编译原理才行啊

wangs 2012-04-14 00:43 发表评论

poj 3243(数论-Baby Step, Giant Step��法)

wangs — Thu, 12 Apr 2012 15:14:00 GMT

poj 3243 hdu 2815 baby_step gaint_step ��法

2009-10-01 20:53

baby_step gaint_step ��法基本思想:

对于一个n个元素的循环(n很大很大) 先算出前面m�?baby_step) 然后以m��?gaint_step)大蟩那么跳了n/m步以�?一定能跛_��前面��出来的m步里�?�q�样旉��复杂度就降到O(m+n/m) �I�间复杂度�ؓ(f��)O(m)

对于计算a^x==b(mod n)中的x

先计��b,b*a,b*a^2,...b*a^m 然后计算1�Q�a^m,a^2m,a^3m,... 那么�l�过i�?��是��C��a^(i*m)的时�?发现它等于b*a^j 那么x=i*m-j

一般m定�ؓ(f��)sqrt(n)�q��时空(�q�且�q�样旉��复杂度最低）(j��) 查找用hash 事实证明map是非常慢�?/p>

//更新

�l�过AekdyCoin盖世��牛的检�?我那个能在poj上跑的程序在hdu上先MLE 然后TLE 然后CE 然后RE 然后WA 千辛万苦最后蟩�q�PE 变成AC�?/p>

原因:动态链表hash跑太�?以后要改成前向星�?/p>

//�l�箋更新

�l�过AekdyCoin教导发现�q�个��法当a和n不互质的时候会(x��)�?因�ؓ(f��)没有逆元 i*m-j不能随便�?/p>

于是�q�夜开发不互质��法如下

设某质数p在a里的指数是ap 在n里面是np 在b里面是bp

那么当x很大 ap*x必然大于np �q�个时候bp必须不小于np 光��命题也成立

同时 ��n里面的p全部除掉剩下的由于和p互质所以左边a�Q�b可不必除反正最后a^x-b一定整除n

所�?先判断a和n的公��p��因数里面有没有b比n��的若小必死否则直接��n除的和a互质再做完破

注意到每个质数的指数肯定不超�q?0 那么当x大于40以上�Ҏ(gu��)��必然成立当x��的时�?虽然�l�证明也可以化�ؓ(f��)abn互质情况但是不如直接验证所以不��了

写了一天了�Q�先是被qsort()写成qsrot()�l�运行错误了一天，晚上发现错误了算法又��Z��问题。哎哎，不行呀。周末的华�ؓ(f��)�Q�赶紧的�Q�刷几个题再��_(d��)��Q�！
压力山大哈！呵呵�Q�心态最重要�Q�没事没事，开朗豁辑ְ�行�?br />

wangs 2012-04-12 23:14 发表评论

wangs — Wed, 11 Apr 2012 16:18:00 GMT

http://poj.org/problem?id=1061

刚刚学了�Ҏ(gu��)��论，也好几天没有写代码了�Q�生疏了不少�Q��L��有错误，�~�译个程序怎么都那么纠�l�啊�Q�看来水�q�_��在是差啊�Q�！
�U�性同余，方程�q�是很好��构造出来了�Q�extended_euclid�Q�中国剩余定理，可是后面却不指导怎么求最��的整数了：(x��)
看了看大牛的题解�Q�原来这样啊�Q�自己数��得多学学才行�Q�多��x��啊：(x��)

分析:��N��蛙蟩了k��?那么��有(x+mk)-(y+nk)=p*L.

即x-y+(m-n)k=p*L,�?m-n)*k≡(y-x) (mod L).

�q�个�U�性同余方�E�有解当且仅�?span>gcd(m-n,L)|(y-x).

令a=m-n,b=L,c=y-x.用扩展欧几里得解方程ax+by=c.

可以求出原方�E�的一个解.如何求最��正整数解呢?

假设我们已经得到一个x0,令d=gcd(m-n,L),

那么所有解可以表示为x=x0+k*L/d.

设L'=L/d.

Xmin=(x0 mod L'+L') mod L'.

WA两次�Q?Ms�Q�囧�Q�还有一�ơ编译错误！�Q�！

#include<stdio.h>
#include<string.h>
#include<math.h>
long long  c,d;
long long gcd_ext(long long a,long long b)
{
    long long gcd,t;
    if (!b)
    {
        c=1;d=0;
        return a;
    }
    gcd=gcd_ext(b,a%b);
    t=c;c=d;d=t-a/b*d;
    return gcd;
}
int main()
{
    long long x,y,m,n,L,a,b,gcd;
    while (scanf("%I64d%I64d",&x,&y)==2)
    {
        scanf("%I64d%I64d%I64d",&m,&n,&L);
        a=m>n?m-n:n-m;
        b=m>n?y-x:x-y;
        gcd=gcd_ext(a,L);
        L=L/gcd;
        if (b%gcd==0)
            printf("%I64d\n",((c*b/gcd)%L+L)%L);
        else
            printf("Impossible\n");
    }
    return 0;
}

�ȝ��Q�代码，�q�是得天天写�Q�三日不�l�手生。自己多��x��Q�多思考思考才能提升能力哈�?/p>

不要��L��ȝ��别�h的题解，要有自己的思�\哈�?/p>

数论�Q�还得��l�看�Q��l�学。要吃透才行�?br />

wangs 2012-04-12 00:18 发表评论

pi的计��，四行c代码�Q�！�Q�orz

wangs — Thu, 05 Apr 2012 07:04:00 GMT

Orz Orz

#include<stdio.h>
#include<string.h>
#include<math.h>
int a=10000,b,c=8400,d,e,f[8401],g;
main()
{
    for(; b-c;)
        f[b++]=a/5;
    for(; d=0,g=c*2; c-=14,printf("%.4d",e+d/a),e=d%a)
        for(b=c; d+=f[b]*a,f[b]=d%--g,d/=g--,--b; d*=b);
}

到底是什么算法呢�Q�得研究研究�Q?img src ="http://www.shnenglu.com/ArcTan/aggbug/170161.html" width = "1" height = "1" />

wangs 2012-04-05 15:04 发表评论

hdu-3666(差分�U�束�pȝ��)

wangs — Thu, 05 Apr 2012 06:42:00 GMT

http://acm.hdu.edu.cn/showproblem.php?pid=3666

2010 Asia Regional Harbin
中的G�?br />
��里推荐做做�q�个�Q�看了看�Q�不知道怎么建图�?br />真s,看了log也还是没有反应过来，
哎哎�Q�log(ai/bj)=log(ai)-log(bj)嘛，�q�样��徏图了啊！�Q�！�Q�！

system of difference constraints�Q?br />WA 16�ơ，气�h啊啊啊啊啊，真心现在也不知道是哪里错了：(x��)

#include<stdio.h>
#include<string.h>
#include<math.h>
#define Max 0xfffffff
int N,M;
int que[1550000],into[805],vis[805];
double    dis[805],map[805][805];

int spfa()
{
    int head,tail,now,i;
    memset(into,0,sizeof(into));
    for (i=1; i<=N ; i++ )
        que[i]=i,vis[i]=0,dis[i]=Max;
    head=0;tail=N;que[0]=0;dis[0]=0.0;
    while (head<=tail)
    {
        now=que[head++];
        vis[now]=1;
        into[now]++;
        if (into[now]>4)
            return 0;
        for (i=1; i<=N ; i++ )
            if (dis[now]+map[now][i]<dis[i])
            {
                dis[i]=dis[now]+map[now][i];
                if (vis[i])
                    que[++tail]=i,vis[i]=0;
            }
    }
    return 1;
}
int main()
{
    int i,j;
    double L,U,a;
    while (scanf("%d%d%lf%lf",&N,&M,&L,&U)==4)
    {
        for (i=0; i<=N+M ; i++ )
            for (j=0; j<=N+M ; j++ )
                map[i][j]=Max;
        for (i=0; i<=N+M ; i++ )
            map[0][i]=0.0;
        U=log(U);L=log(L);
        for (i=1; i<=N ; i++ )
            for (j=1; j<=M ; j++ )
            {
                scanf("%lf",&a);
                map[j+N][i]=U-log(a);
                map[i][j+N]=log(a)-L;
            }
        N=N+M;
        puts(spfa()?"YES":"NO");
    }
    return 0;
}

差分�U�束�pȝ��Q�是�U�性规划的一�U�特例。得研究研究它的对偶问题是什么，嘿嘿�Q�好东西呀�Q�！
建立模型很重要哇�Q�！

图论的最短�\�Q�好多东西呢�Q�得好好学学啊，spfa是个好东�ѝ��以后要多学��法多看论文了！
spfa�Q�可以判断负权回路哈。这个题目比较弱�Q?�ơ就可以了。夜游sqrt(|v|)的，n当然是一个上界�?img src ="http://www.shnenglu.com/ArcTan/aggbug/170155.html" width = "1" height = "1" />

wangs 2012-04-05 14:42 发表评论

wangs — Thu, 05 Apr 2012 06:35:00 GMT

LinerProgramming�Q�线性规划。是�q�筹学的一个重要分支�?br />1947�q�单��h��Q�G.B.Dantzing�Q�提��Z��一般LP规划问题的求解方�?#8212;——单纯型法�Q�simplex algorithm�Q��?br />�q�里是用到没有改�q�的单纯型法�Q�输入数据的标准型^..^

该学学改�q�的单纯型法了�?br />
�U�性规划问题还是有很多用的�Q�是个好模型�Q�！�Q?br />

#include<stdio.h>
#include<string.h>
#include<math.h>
#define inf 10000000.0
int n,m;
int g[1005],q[1005],p[1005];
double b[1005],c[1005],x[1005],a[1005][1005];

//********数据输入、初始化*************
int init()
{
    int i,j;
//********数组初始�?*********
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    memset(c,0,sizeof(c));
    memset(x,0,sizeof(x));
    memset(q,0,sizeof(q));
    printf("误��入单�U�型表的标准�?\n");
//********输入数据************
    scanf("%d%d",&m,&n);
    for (i=1; i<=n ; i++ )
        scanf("%lf",&c[i]);
    for (i=1; i<=n ; i++ )
        scanf("%d",&p[i]);
    for (i=1; i<=m ; i++ )
    {
        for (j=1; j<=n ; j++ )
            scanf("%lf",&a[i][j]);
        scanf("%lf",&b[i]);
    }
//**********初始化单�U�型�?***
    for (i=1; i<=m ; i++ )
        g[i]=i,q[i]=1,a[i][n+1]=b[i],x[i]=b[i];
    for (j=1; j<=n ; j++ )
        a[m+1][j]=c[j];
    for (j=n+1; j>m ; j-- )
    {
        for (i=1; i<=m ; i++ )
            a[m+1][j]-=c[i]*a[i][j];
    }
}
//**********�l�果输出********************
int print(int result)
{
    int i;
    double sum;
    if (result==-1)
    {
        printf("无可行解\n");
        return ;
    }
    if (result==-2)
    {
        printf("无界解\n");
        return ;
    }
    if (result==-3)
    {
        printf("无穷多最优解。其中一个是�Q�\n");
        sum=0.0;
        for (i=1; i<=n ; i++ )
            sum+=x[i]*c[i];
        for (i=1; i<n ; i++ )
            printf("%.4lf ",x[i]);
        printf("%.4lf\n",x[n]);
        return ;
    }
    printf("有最优解�Q�\n");
    sum=0.0;
    for (i=1; i<=n ; i++ )
        sum+=x[i]*c[i];
    printf("%.4lf\n",sum);
    for (i=1; i<n ; i++ )
        printf("%.4lf ",x[i]);
    printf("%.4lf\n\n",x[n]);
    return ;
}
//***********��(g��)查单�U�型�?**************
int check()
{
    int i,j,flag,flg,mj;
    double max;
    flag=0;
    flg=0;
    for (j=1; j<=n ; j++ )
    {
        if (a[m+1][j]>0.0)
        {
            flag=1;
            max=0.0;
            for (i=1; i<=m ; i++ )
                if (a[i][j]>max)
                    max=a[i][j],mj=j;
            if (max>0.0)
                flg=1;
        }
    }
    if (!flag)
    {
        for (i=1; i<=m ; i++ ) //判断是否无可行解
        {
            if (p[g[i]]&&a[i][n+1]!=0.0)
                return -1;
        }

        for (j=1; j<=n ; j++ ) //判断是否有无�I�多最优解
            if (!q[j]&&a[m+1][j]==0.0)
                return -3;
        return 0;//唯一最优解
    }
    if (!flg)
        return -2;//无界�?/span>

    return mj;//扑ֈ�最大的那个a[m+1][j]作�ؓ(f��)换入变量
}
//*********扑ֈ�最��的那个�?************��L��能找到的�Q�？�Q�？
int f_min(int r)
{
    int i,mi;
    double min;
    min=inf;
    for (i=1; i<=m ; i++ )
        if (a[i][r]!=0.0&&a[i][n+1]/a[i][r]>0&&a[i][n+1]/a[i][r]<min)
            min=a[i][n+1]/a[i][r],mi=i;
    printf("%.4lf ",min);
    return mi;//��定为换出变�?/span>
}
//********guass消元法进行�P�?**********
int guass(int k,int r)
{
    int i,j;
    for (j=n+1; j>=1 ; j-- ) //行变�?/span>
        if (j!=r)
            a[k][j]/=a[k][r];
    a[k][r]=1.0;
    for (i=1; i<=m+1 ; i++ ) //每一行进行变�?/span>
        if (i!=k)
        {
            for (j=1; j<=n+1 ; j++ )
                if (j!=r)
                    a[i][j]-=a[k][j]*a[i][r];
            a[i][r]=0.0;
        }
    q[g[k]]=0;
    g[k]=r;
    q[r]=1;
    memset(x,0,sizeof(x));
    for (i=1; i<=m ; i++ )
        x[g[i]]=a[i][n+1];
}
int work()
{
    int i,j,r,k;
    while (1)
    {
        for (i=1; i<=m+1 ; i++ )
        {
            for (j=1; j<=n+1 ; j++ )
                printf("%.4lf ",a[i][j]);
            printf("%d\n",g[i]);
        }
        for (j=1; j<=n ; j++ )
            printf("%.4lf ",x[j]);
        printf("%.4lf\n",x[j]);
        r=check();
        if (r<=0)
            return r;
        k=f_min(r);
        printf("%d %d\n\n",k,r);

        guass(k,r);
    }
}
int main()
{
    int result;
    init();
    result=work();
    print(result);
    return 0;
}

有时间得去poj�Q�zoj上找扄��性规划的题目来下写写。嘿嘿，

wangs 2012-04-05 14:35 发表评论

poj3414(bfs)

wangs — Mon, 02 Apr 2012 09:47:00 GMT

http://poj.org/problem?id=3414

bfs:都是些基本的��法了，各种bfs写了不少啦�?br />�q�个题目也挺��单的�Q�主要是那几个�{�U�L��向还是要优化下好写，觉得自己写得�q�可以哈�Q?br />嘿嘿�Q?br />bfs:0ms

#include<stdio.h>
#include<string.h>
#include<math.h>
int a[2],c,head,tail;
int que[10005][2],pre[10005],p[10005],vis[105][105],ans[10005];
int check()
{
    if (que[tail][0]==c||que[tail][1]==c)
        return 1;
    return 0;
}
int Fill(int i)
{
    int t=tail+1;
    que[t][0]=que[head][0];
    que[t][1]=que[head][1];
    que[t][i-1]=a[i-1];
    if (!vis[que[t][0]][que[t][1]])
    {
        vis[que[t][0]][que[t][1]]=1;
        p[t]=i;
        pre[t]=head;
        return 1;
    }
    return 0;
}
int Drop(int i)
{
    int t=tail+1;
    que[t][0]=que[head][0];
    que[t][1]=que[head][1];
    que[t][i-1]=0;
    if (!vis[que[t][0]][que[t][1]])
    {
        vis[que[t][0]][que[t][1]]=1;
        p[t]=2+i;
        pre[t]=head;
        return 1;
    }
    return 0;
}
int Four(int i,int j)
{
    int min,t=tail+1;
    min=(que[head][i-1])<(a[j-1]-que[head][j-1])?(que[head][i-1]):(a[j-1]-que[head][j-1]);
    que[t][j-1]=que[head][j-1]+min;
    que[t][i-1]=que[head][i-1]-min;
    if (!vis[que[t][0]][que[t][1]])
    {
        vis[que[t][0]][que[t][1]]=1;
        p[t]=4+i;
        pre[t]=head;
        return 1;
    }
    return 0;
}
int bfs()
{
    int i,tmp;
    memset(vis,0,sizeof(vis));
    head=0;
    tail=1;
    que[1][0]=0;
    que[1][1]=0;
    vis[0][0]=1;
    pre[1]=0;
    p[1]=0;
    while (head<tail)
    {
        head++;
        tmp=1;
        for (i=1; i<=2 ; i++ )
        {
            if (Fill(i))
                tail++;
            if (check())
                return tail;
            if (Drop(i))
                tail++;
            if (check())
                return tail;
            if (Four(i,i+tmp))
                tail++;
            if (check())
                return tail;
            tmp=-1;
        }
    }
    return 0;
}
int print(int k)
{
    int i;
    if (!k)
    {
        printf("impossible\n");
        return 0;
    }
    i=0;
    while (k)
    {
        ans[++i]=p[k];
        k=pre[k];
    }
    printf("%d\n",i-1);
    for (k=i-1; k>=1 ; k-- )
    {
        if (ans[k]<=2)
        {
            printf("FILL(%d)\n",ans[k]);
            continue;
        }
        if (ans[k]<=4)
        {
            printf("DROP(%d)\n",ans[k]-2);
            continue;
        }
        if (ans[k]==5)
            printf("POUR(1,2)\n");
        else
            printf("POUR(2,1)\n");
    }
    return 0;
}

int main()
{
    int k;
    scanf("%d%d%d",&a[0],&a[1],&c);
    if (c>a[0]&&c>a[1])
        k=0;
    else
        k=bfs();
    print(k);
    return 0;
}

�q�几天的bfs�Q�dfs�Q�都是第一�ơ就AV�Q�挺爽的�Q�嘿�ѝ�?br />多写写恶心的代码�Q�以后就不恶心了�Q�哈哈�?br />
很纳��P��那些用很��内存的是怎么做到的呢�Q�看来自��p��是很弱啊～～�?br />

wangs 2012-04-02 17:47 发表评论

poj1419(最大团、最大独立集)

wangs — Mon, 02 Apr 2012 07:14:00 GMT

http://poj.org/problem?id=1419

昄��的最大独立集�Q�我晕啊�Q�最大团没有学好啊，没有真正理解清楚
哎哎
模版题，最大独立集�Q?br />

#include<stdio.h>
#include<string.h>
#include<math.h>
int map[100][100],x[100],b[100];
int n,m,max,now;
int dfs(int i)
{
    int j,flag;
    if (i>n)
    {
        for (j=1; j<=n ; j++ )    x[j]=b[j];
        max=now;
        return 0;
    }
    flag=1;
    for (j=1; j<i ; j++ )
        if (b[j]&&map[i][j])   //�q�里如果改�ؓ(f��)�Q�map[i][j]��是求最大团了。。。�?br />        {
            flag=0;
            break;
        }
    if (flag)
    {
        b[i]=1;
        now++;
        dfs(i+1);
        now--;
        b[i]=0;
    }
    if (now+n-i>max)
    {
        b[i]=0;
        dfs(i+1);
    }
}
int main()
{
    int i,j,t,x1,x2,flag;
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d%d",&n,&m);
        memset(map,0,sizeof(map));
        for (i=1; i<=m ; i++ )
        {
            scanf("%d%d",&x1,&x2);
            map[x2][x1]=map[x1][x2]=1;
        }

        max=0;now=0;
        dfs(1);
        printf("%d\n",max);
        flag=0;
        for (i=1; i<=n ; i++ )
            if (x[i])
            {
                if (flag)
                    printf(" ");
                printf("%d",i);
                flag=1;
            }
        printf("\n");
    }
    return 0;
}

wangs 2012-04-02 15:14 发表评论

poj1426(bfs、dp)

wangs — Mon, 02 Apr 2012 06:16:00 GMT

http://poj.org/problem?id=1426

bfs:刚刚拿到��x��所有数�l�用字符串保存，�q�样最多可�?00�?1呢�?br /> 一直都觉得�q�样朴素的算法很恶心�Q�就��d��饭去了，我嚓�Q?2点吃早饭�?br /> �H�然惛_��其实我的串可以que[]里存0�?嘛，然后索引一下就好了于是乎，行了�Q�嘿�ѝ��这样递推�q�去的，其实应该是可以整��Z��dp状态�{�U�L��E�的哈，�{�会(x��)��x��?br /> bfs:
32MS

#include<stdio.h>
#include<string.h>
#include<math.h>
int n;
int que[1000005],pre[1000005],q[1000005],ans[205];//que[] 0�?�Q�q[]当前余数�Q�pre[]前驱�?br />int bfs()
{
    int head,tail,now,next,i;
    head=0;tail=1;que[1]=1;pre[1]=0;q[1]=1;
    while (head<tail)
    {
        now=q[++head]*10;
        for (i=0;i<=1 ;i++ )
        {
            next=now+i;
            tail++;
            que[tail]=i;
            q[tail]=next%n;
            pre[tail]=head;
            if (!q[tail])
                return tail;
        }
    }
}
int main()
{
    int i,k;
    while (scanf("%d",&n)==1&&n)
    {
        i=bfs();
        k=0;
        while (i)
        {
            ans[++k]=que[i];
            i=pre[i];
        }
        for (i=k;i>=1 ;i-- )
            printf("%d",ans[i]);
        printf("\n");
    }
    return 0;
}

同学用的long long 也过了，�Ҏ(gu��)��研究研究哈�?br />long long:我晕�Q�居然跑�?10MS

dp�?x��)跑多久么�?br />

wangs 2012-04-02 14:16 发表评论

wangs — Mon, 02 Apr 2012 03:27:00 GMT

171A - Mysterious numbers - 1

The easiest way to make the problem statement unusual is to omit it. This is an extremely convenient approach — you don’t have to maintain the statement in two languages or to worry that it might turn out to be ambiguous or too long or too scary. 690 people solved this problem, so evidently we can omit statements even in regular rounds :-)

As for the problem itself, it required to sum the first number and the reverse of the second number.

171B - A star

They say it’s better to see once than to hear ten times or to read a hundred times. In this problem we decided to check this and to replace the traditional textual statement with a single image. Same as in the previous problem, it did well — at least 645 participants recognized star numbers (sequence http://oeis.org/A003154 in OEIS), the numbers of balls needed to form a six-pointed start of certain size. After this one had only to code the formula — S_n = 6n(n−1) + 1.

171D - Broken checker

What does one do if the statement is unknown and the only source of information about the problem is the checker? Right — you just try all possible functions which convert 5 input values into 3 output values and see which of them fits :-)

171F - ucyhf

This problem finally has a statement! The trick is, it’s encoded. We decided to be kind to you and to use the simplest cipher possible — Caesar cipher (each letter is shifted the same number of positions in the alphabet). By a long stretch of imagination one could break the cipher by hand — observe frequent letters and short words, deduce possible values of shift and verify it against the rest of the message. A lazier one could Goog

le for a tool like this one and get the decoded version semi-automatically.

After cracking the statement the rest was almost easy — you had to find a prime whose reverse is a prime different from the original one (sequence http://oeis.org/A006567). 11184-th such number equals 999983, so one could do a brute-force check of all numbers in row.

171E - UNKNOWN LANGUAGE

A special contest written by me and no special programming language? Impossible! I pulled myself up and made only 25% of all problems esoteric — that’s two. There really should be three but the third interpreter refused to cooperate. Maybe next time…

What can one do if all he knows about the language is its compiler? Just run any code and see what the compiler says. In this case the compiler said “DO YOU EXPECT ME TO FIGURE THIS OUT?”, and Google should tell you immediately that the language in question is INTERCAL. The problem simplifies to figure out the dialect used and how to output “INTERCAL” in it.

In Codeforces round #96 I gave a problem 133C - Turing Tape which explained the mechanism of string output in INTERCAL and asked to write a program which converted a string into an array of numbers which would print this string. Combine this knowledge with Hello, World! example and you get the program you need. Actually, that’s what I did to write the reference solution.

PLEASE DO ,1 <- #8 DO ,1 SUB #1 <- #110 PLEASE DO ,1 SUB #2 <- #32 DO ,1 SUB #3 <- #72 DO ,1 SUB #4 <- #136 DO ,1 SUB #5 <- #88 DO ,1 SUB #6 <- #136 DO ,1 SUB #7 <- #64 DO ,1 SUB #8 <- #80 PLEASE READ OUT ,1 PLEASE GIVE UP

171C - A Piece of Cake

The second esoteric problem had a Chef program as the statement. You had only to figure out what it does and do it in any regular language. It turns out that this program reads N followed by N numbers $a_1, a_2, …, a_N$ and calculated the sum i * a_i.

171G - Mysterious numbers - 2

This one was much harder to guess but much easier to code. First two numbers were the start of a Fibonacci-like sequence, and the third one was the index of the required number in this sequence.

�W�一�ơ做CF�Q�看来是没有选好旉��啊，呵呵�Q�April Fools Day�Q�把我娱乐愚了�?br />��主说这是个智慧的比赛，我承认，自己智商不行不行~~~~
A�Q?br />

B�Q?br />

C�Q?br />

D�Q?br />

G�Q?br />

E和F表示不会(x��)啊，再研�I�研�I�吧。弱爆了~~~~

��主��实很厉宛_��Q�看来今�q�又果断�?x��)被虐了�?br />
惌��力不行呀�Q?img src ="http://www.shnenglu.com/ArcTan/aggbug/169845.html" width = "1" height = "1" />

wangs 2012-04-02 11:27 发表评论