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cometrue — Mon, 06 Dec 2010 10:22:00 GMT

1 #include <iostream>
2 #include <cstring>
3 using namespace std;
4 char a[36000];
5 void rev()
6 {
7     int len=strlen(a),i;
8     char t;
9     for(i=0;i<len/2;++i)
10     {
11         t=a[i];
12         a[i]=a[len-1-i];
13         a[len-1-i]=t;
14     }
15 }//strrev()貌似不是标准库函敎ͼ��?br>16
17 void multi(int n)
18 {
19     int i,l=strlen(a),m=0,jw=0;
20     rev();
21     char t[36000];
22     for(i=0;i<l;++i)
23     {
24         t[i]=((a[i]-'0')*n+jw)%10+'0';
25         jw=((a[i]-'0')*n+jw)/10;
26     }
27     if(jw>=1000)
28     {
29         t[i]=jw%10+'0';
30         t[i+1]=(jw/10)%10+'0';
31         t[i+2]=(jw/100)%10+'0';
32         t[i+3]=jw/1000+'0';
33         t[i+4]='\0';
34     }
35     else if(jw>=100)
36     {
37         t[i]=jw%10+'0';
38         t[i+1]=(jw/10)%10+'0';
39         t[i+2]=jw/100+'0';
40         t[i+3]='\0';
41     }
42     else if(jw>=10)
43     {
44         t[i]=jw%10+'0';
45         t[i+1]=(jw/10)%10+'0';
46         t[i+2]='\0';
47     }
48     else if(jw)
49     {
50         t[i]=jw+'0';
51         t[i+1]='\0';
52     }
53     else t[i]='\0';
54     strcpy(a,t);
55     rev();
56 }//��字�W�串乘n�Q�需考虑最后的�q�位的位数�?br>57
58 int main()
59 {
60     int n;
61     while(cin>>n)
62     {
63         memset(a,0,36000);
64         a[0]='1';
65         a[1]='\0';
66         for(int i=2;i<=n;++i)multi(i);
67         cout<<a<<endl;
68     }
69     return 0;
70 }
71

�׃��一直不肯写个大整数的类�Q�又不会(x��)用JAVA�Q�遇到这�U�题目真是感到很隑֏�。不�q�我今天用了一�U�比较耗时但确实思�\��单的�Ҏ(gu��)��q�了�q�道题。首先，我们必须知道10000�Q�到底有多少位，�q�样才好定义合适的数组�?/div>

log10�Q?�Q?log(3)+...+log10�Q?0000�Q?35659.9�Q�所以定义一�?6000的字�W�数�l�就够了。整个实现比较简单但是用�?312MS.....应该分治之类的算法会(x��)好点�Q�最快的100MS��p��了。估计是重复的反转和复制耗时了�?/div>

cometrue 2010-12-06 18:22 发表评论

cometrue — Wed, 24 Nov 2010 05:35:00 GMT

//求N!的位�?br>
//N!=1*2*3**N�Q�两边取常用�Ҏ(gu��)��Q�即可算出log10(N!),向上取整即�ؓ(f��)N!的位�?br>//hdoj    984MS    344K
/*
#include
#include
using namespace std;
int main()
{
    double sum=0.0;
    int n,i,times,res;
    if(cin>>times&×!=0)
    {
        while(times)
        {
            cin>>n;
            for(i=2;i<=n;++i)
                sum+=log10(i);
            res=ceil(sum);
            cout<            sum=0.0;
            --times;
        }
    }
    return 0;
}
*/
//String公式的方法，N!~sqrt(2*pi*N)*(N/e)^N
//hdoj    0MS        360K
#include <iostream>
#include <cmath>
using namespace std;
const double pi=3.1415926;
int main()
{
    int n,times;
    long double sum;
    if(cin>>times&&times)
    {
        while(times)
        {
            cin>>n;
            sum=(long double)0.5*log10(2*pi*n)+(long double)n*(log10(n)-log10(exp(1)));
            cout<<(long)ceil(sum)<<endl;
            --times;
        }
    }
    return 0;
}

cometrue 2010-11-24 13:35 发表评论

hdoj_1005_Number Sequence

cometrue — Thu, 18 Nov 2010 09:00:00 GMT

#include <iostream>
using namespace std;
int a,b,s[100];
struct Pair
{
    int x;
    int y;
}res[50];
int main()
{
    int n,i,j,k;
    bool flag=false;
    res[0].x=res[0].y=1;
    while(cin>>a>>b>>n)
    {
        if(!(a||b||n))return 0;
        for(i=1;i<50;++i)
        {
            res[i].x=res[i-1].y;
            res[i].y=(a*res[i-1].y+b*res[i-1].x)%7;
            for(j=0;j<i-1;++j)//…………………………注意�q�里循环上限是i-1�Q�这样可以排除三个连�l�相�{�的情况。就是把循环节�ؓ(f��)1的看�?.
            {
                if(res[j].x==res[i].x&&res[j].y==res[i].y)
                {
                    flag=true;
                    break;
                }
            }
            if(flag)break;
        }//一个��@环找出��@环节大小
        flag=false;//……………………注意把标志还�?/span>
        if(n<=j)cout<<res[n].x<<endl;//未进入��@环时
        else
        {
            if((n-j)%(i-j)==0)k=i-1;
            else k=(n-j)%(i-j)+j-1;//�q�个式子改了很长旉��Q��L��?x��)出现问题。这是最�l�的形式
            cout<<res[k].x<<endl;
        }
    }
    return 0;
}

提交了七�ơ终于给�q�了。是道数论的��单题�Q�不�q�应该用不到什么高��q��知识�Q�关键是扑և�循环节。因为对�?000000000的大��，如果不找规律的话无论如何也要��时的。分析一下，每个��C��取决于它前面的两个，所以如果出��C��相同的数对，则必出现循环。而且�Q�每个数都是0~6之间的一个，可知不同的数对只�?*7=49个，那么只要计算出前50个数�Q�则其中必有相同的两�Ҏ(gu��)��出现。上代码。AC之后我想知道循环是不是��L��从最前面两个数开始，于是��单写了一个程序，遍历了所有的a,b(易知它们也只�?9�U�组�?�Q�下面是我得到的�l�果�Q?div>

a b j i i-j

0 0 2 4 2

0 1 0 2 2

0 2 0 6 6

0 3 0 12 12

0 4 0 6 6

0 5 0 12 12

0 6 0 4 4

1 0 0 2 2

1 1 0 16 16

1 2 0 6 6

1 3 0 24 24

1 4 0 48 48

1 5 0 21 21

1 6 0 6 6

2 0 1 4 3

2 1 0 6 6

2 2 0 48 48

2 3 0 6 6

2 4 0 48 48

2 5 0 24 24

2 6 0 2 2

3 0 1 7 6

3 1 0 16 16

3 2 0 48 48

3 3 0 42 42

3 4 0 6 6

3 5 0 2 2

3 6 0 8 8

4 0 1 4 3

4 1 0 16 16

4 2 0 48 48

4 3 0 21 21

4 4 0 2 2

4 5 0 6 6

4 6 0 8 8

5 0 1 7 6

5 1 0 6 6

5 2 0 48 48

5 3 0 2 2

5 4 0 48 48

5 5 0 24 24

5 6 0 14 14

6 0 1 3 2

6 1 0 16 16

6 2 0 2 2

6 3 0 24 24

6 4 0 48 48

6 5 0 42 42

6 6 0 3 3

可见当a,b都是7的倍数�Ӟ��循环从第三个数开始（以后都是0�Q�；当a,b中只有一个是7的倍数�Ӟ��循环从第二个数开始（1,0�?,1的情冉|��较特�D�，因�ؓ(f��)跟开始的1,1重复了所以可以认为是从第一个数开始）�Q�当a,b都不�?的倍数是，循环从第一个数开始。可见还是从�W�一个数开始��@环的多。��@环节也有长有短，比如当a=1,b=4时一直到�W?9个数才出现��@环�?/div>

cometrue 2010-11-18 17:00 发表评论

USACO_Section 1.2_palsquare

cometrue — Thu, 21 Oct 2010 09:32:00 GMT

#include <stdio.h>
#include <string.h>
void conv(char numb[],int n,int base)
{
    int num[18],len=0,j;
    while(n/base)
    {
        num[len]=n%base;
        ++len;
        n/=base;
    }
    num[len]=n;


    for(j=len;j>=0;--j)
    {
        if(num[j]>9)numb[len-j]=num[j]+55;
        else numb[len-j]=num[j]+'0';
    }
    numb[len+1]='\0';
    return ;
}

int main()
{
    FILE *fin,*fout;
    fin=fopen("palsquare.in","r");
    fout=fopen("palsquare.out","w");
    int base,i,len=0,j;
    fscanf(fin,"%d",&base);
    for(i=1;i<=300;++i)
    {
        char square[18]={'\0'},num[10]={'\0'};
        int flag=1;
        conv(num,i,base);
        conv(square,i*i,base);
        len=strlen(square);
        for(j=0;j<=len/2;++j)
        {
            if(square[j]!=square[len-j-1])
            {
                flag=0;
                break;
            }
        }
        if(flag)fprintf(fout,"%s %s\n",num,square);
    }
    return 0;
}

我还是习(f��n)惯用C�?#8230;…所以把代码贴上来的时候发现stdio是黑色的�Q��?#8220;base”是蓝色的�?/div>

��p��样吧�?/div>

题目�Q?/div>

Palindromic Squares
Rob Kolstad

Palindromes are numbers that read the same forwards as backwards. The number 12321 is a typical palindrome.

Given a number base B (2 <= B <= 20 base 10), print all the integers N (1 <= N <= 300 base 10) such that the square of N is palindromic when expressed in base B; also print the value of that palindromic square. Use the letters 'A', 'B', and so on to represent the digits 10, 11, and so on.

Print both the number and its square in base B.

PROGRAM NAME: palsquare

INPUT FORMAT

A single line with B, the base (specified in base 10).

SAMPLE INPUT (file palsquare.in)

OUTPUT FORMAT

Lines with two integers represented in base B. The first integer is the number whose square is palindromic; the second integer is the square itself.

SAMPLE OUTPUT (file palsquare.out)

没有什么复杂的��法�Q�因��一节讲的就�?#8220;the brute force, straight-forward, try-them-all method of finding the answer.

”

cometrue 2010-10-21 17:32 发表评论

USACO_Section 1.1_beads

cometrue — Thu, 21 Oct 2010 06:54:00 GMT

#include <stdio.h>
#include <stdlib.h>
int main()
{
    FILE *fin,*fout;
    fin=fopen("beads.in","r");
    fout=fopen("beads.out","w");
    char *beads;
    int n;
    fscanf(fin,"%d",&n);
    beads=(char *)malloc(3*n*sizeof(char));
    fscanf(fin,"%s",beads);
    int i,a,b,left,right,sum=0;
    for(i=n;i<3*n;++i)
    {
        beads[i]=beads[i-n];
    }
    for(i=n;i<2*n;++i)
    {
        left=i;
        right=i+1;
        char ch;

        while(beads[left]=='w'&&left>=0)--left;
        ch=beads[left];
        while(left>0&&(beads[left-1]==ch||beads[left-1]=='w'))--left;
        a=i-left+1;

        while(beads[right]=='w'&&right<3*n)++right;
        ch=beads[right];
        while(right<(3*n-1)&&(beads[right+1]==ch||beads[right+1]=='w'))++right;
        b=right-i;

        if(a+b>sum)sum=a+b;
        if(a>=n||b>=n||a+b>n)sum=n;
    }
    fprintf(fout,"%d\n",sum);
    return 0;
}

首先我的��x��是从1到n,left=0,right=1,然后往两边数颜色相同的珠子。如果用一个大��ؓ(f��)n的数�l�存字符�Ԍ��一个很昄��的问题就是当left<0或者right>n-1时就要溢出。所以要用到一个取余的函数�?/div>

但是�q�样��实太麻烦了�Q�写的代码也�Ҏ(gu��)��出错�Q�我�l�于军_��重写了。新的想法是在字�W�串两边各复制一份相同的�Q�这样就是大��ؓ(f��)3×n的字�W�串�Q�而��@环时只需要从n�?×n-1,解决了溢出的问题。（但是我觉得这�q�不是一个好�Ҏ(gu��)��Q�因为浪费了三倍的�I�间�Q�。最�l�的代码是这��L(f��ng)��Q�虽然AC了，但��M��是那么完��?/div>

cometrue 2010-10-21 14:54 发表评论

USACO_Section 1.1_beads

cometrue — Thu, 21 Oct 2010 06:39:00 GMT

题目不难�Q�但是。。�?div>首先我的��x��是从1到n,left=0,right=1,然后往两边数颜色相同的珠子。如果用一个大��ؓ(f��)n的数�l�存字符�Ԍ��一个很昄��的问题就是当left<0或者right>n-1时就要溢出。所以要用到一个取余的函数

int cycle(int a,int n)

{

return a<0?(a%n+n):(a%n);

}

#include

int main()

{

FILE *fin,*fout;

fin=fopen("beads.in","r");

fout=fopen("beads.out","w");

char *beads;

int n;

fscanf(fin,"%d",&n);

beads=(char *)malloc(3*n*sizeof(char));

fscanf(fin,"%s",beads);

int i,a,b,left,right,sum=0;

for(i=n;i<3*n;++i)

{

beads[i]=beads[i-n];

}

for(i=n;i<2*n;++i)

{

left=i;

right=i+1;

char ch;

while(beads[left]=='w'&&left>=0)--left;

ch=beads[left];

while(left>0&&(beads[left-1]==ch||beads[left-1]=='w'))--left;

a=i-left+1;

while(beads[right]=='w'&&right<3*n)++right;

ch=beads[right];

while(right<(3*n-1)&&(beads[right+1]==ch||beads[right+1]=='w'))++right;

b=right-i;

if(a+b>sum)sum=a+b;

if(a>=n||b>=n||a+b>n)sum=n;

}

fprintf(fout,"%d\n",sum);

return 0;

}

cometrue 2010-10-21 14:39 发表评论