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HDOJ 1205 HDU 1205 吃糖�?ACM 1205 IN HDU

MiYu — Thu, 12 Aug 2010 02:41:00 GMT

MiYu原创, 转帖��h��?: 转蝲�?______________白白の屋

题目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=1205
题目描述:

吃糖�?br>
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 2000/1000 K (Java/Others)
Total Submission(s): 4426 Accepted Submission(s): 1310

Problem Description
HOHO�Q�终于从Speakless手上赢走了所有的�p�果�Q�是Gardon吃糖果时有个�Ҏ��的癖好，��是不喜�Ƣ将一��L��p�果攑֜�一起吃�Q�喜�Ƣ先吃一�U�，下一�ơ吃另一�U�，�q�样�Q�可是Gardon不知道是否存在一�U�吃�p�果的顺序��得他能把所有糖果都吃完�Q�请你写个程序帮忙计��一下�?br>

Input
�W�一行有一个整数T�Q�接下来T�l�数据，每组数据�?行，�W�一行是一个整数N�Q?/span>0<N<=1000000)�Q�第二行是N个数�Q�表�C�N�U�糖果的数目Mi(0<Mi<=1000000)�?br>

Output
对于每组数据�Q�输��Z��行，包含一�?/span>"Yes"或�?/span>"No"�?br>

Sample Input
2
3
4 1 1
5
5 4 3 2 1

Sample Output
No
Yes

题目分析 :

先吃一个最多的, 再吃一个其他的,只要最多的满�� : sum - max + 1 >= max , 那么��可以吃�?

代码如下:

main(T,N,num,max){long long sum;scanf("%d",&T);while(T--){scanf("%d",&N);sum=0,max=0;while(N--){scanf("%d",&num);max=num>max?num:max;sum+=num;}sum-=max-1;puts(sum>=max?"Yes":"No");}}

MiYu 2010-08-12 10:41 发表评论

HDOJ HDU 1235 �l�计同成�l�学生�h�?ACM 1235 IN HDU

MiYu — Mon, 09 Aug 2010 01:22:00 GMT

MiYu原创, 转帖��h��?: 转蝲�?______________白白の屋

题目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=1235

很谁的题, 直接 hash ��可以了.
代码如下 :

#include <iostream>
#include <map>
using namespace std;
int main ()
{
    int N;
    while ( scanf ( "%d",&N ) , N )
    {
            map < int ,int > mp;
            int num;
            for ( int i = 0; i != N; ++ i )
            {
                  scanf ( "%d",&num );
                  mp[num] ++;
            }
            scanf ( "%d",&num );
            cout << mp[num] << endl;
    }
    //getchar();
    return 0;
}

main (N,num)
{int scr[101],i;while(scanf("%d",&N),N){
memset ( scr , 0, sizeof ( scr ) );
for (i = 0; i != N; ++ i )
{scanf ( "%d",&num );scr[num] ++;}
scanf ("%d",&num);
printf ("%d\n",scr[num]);}
}

MiYu 2010-08-09 09:22 发表评论

HDOJ HDU 1201 18岁生�?ACM 1201 IN HDU

MiYu — Sun, 08 Aug 2010 14:24:00 GMT

MiYu原创, 转帖��h��?: 转蝲�?______________白白の屋

题目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=1201
题目描述:

Problem Description
Gardon�?8岁生日就要到了，他当然很开心，可是他突然想��C��个问题，是不是每个�h从出生开始，到达18岁生日时所�l�过的天数都是一��L��呢？��g��q�不全都是这��P��所以他惌��你帮忙计��一下他和他的几个朋友从出生到达18岁生日所�l�过的��d��敎ͼ�让他好来比较一下�?br>

Input
一个数T�Q�后面T行每行有一个日期，格式是YYYY-MM-DD。如我的生日�?988-03-07�?br>

Output
T行，每行一个数�Q�表�C�此��Z��出生�?8岁生日所�l�过的天数。如果这个�h没有18岁生日，��p��?/span>-1�?br>

Sample Input
1
1988-03-07

Sample Output
6574

水题, 注意有没�?8岁和瑞年的判断就好了

代码如下 ( 参照奋斗�?) :

#include <iostream>
#include <cmath>
using namespace std;
bool isLeap ( int n )
{
        return !( n % ( n % 100 ? 4 : 400 ) );
}

int main ()
{
       int T;
    scanf ( "%d", &T );
       while ( T -- )
       {
            int year, month, day;
               scanf ( "%d-%d-%d", &year, &month, &day );
                  if ( month == 2 && day == 29 )
                  {
                          cout << -1 << endl;
                          continue;
                  }
                  int run = 0;
                  if ( month >= 3 )
                  {
                          for ( int i = 1; i <= 18; ++ i )
                             {
                       if ( isLeap ( year + i ) )
                       {
                            run ++;
                       }
                 }
                  }
                  else
                  {
                          for ( int i = 0; i < 18; ++ i )
                             {
                       if ( isLeap ( year + i ) )
                       {
                                           run ++;
                       }
                 }
                  }
                  cout << 365 * 18 + run << endl;
       }
       return 0;
}

MiYu 2010-08-08 22:24 发表评论

HDOJ HDU 1202 The calculation of GPA ACM 202 IN HDU

MiYu — Sun, 08 Aug 2010 12:27:00 GMT

MiYu原创, 转帖��h��?: 转蝲�?______________白白の屋

题目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=1202
题目描述:

Problem Description
每学期的期末�Q�大安��会忙于计��自��q��q�_��成�W�Q�这个成�l�对于评奖学金是直接有关的。国外大学都是计��GPA(grade point average) 又称GPR(grade point ratio)�Q�即成�W�Ҏ��与学分的加权�q�_��值来代表一个学生的成�W的。那么如何来计算GPA呢？

一般大学采用之计分�?nbsp;

A90 - 100 4 �?nbsp;
B80 - 89 3 �?nbsp;
C70 - 79 2 �?nbsp;
D60 - 69 1 �?nbsp;
E0 - 59 0 �?nbsp;

例如�Q�某位学生修习三门课�Q�其评��、学分及成�W分别为：
英文�Q�三学分�?/span>92 分；化学�Q�五学分�?/span>80 分；数学�Q�二学分�?0分，则GPA的算法如下：

�U�目学分分数 �Ҏ�� 分数×�Ҏ��
英文  3    92    4     12
化学  5    80    3     15
数学  2    60    1      2
合计  10   29
29/10=2.9
2.9即�ؓ某生的GPA
下面有请你写一个用于计��GPA的程序�?br>

Input
包含多组数据�Q�每�l�数据的�W�一行有一个数N�Q�接下来N行每行表�C�Z��门成�l�。每行有两个实型的数 s,p�Q�s表示�q�门评��学分,p表示该学生的成�W�Q�百分制�Q�。如果p=-1则说明该学生�q�门评��考，是不应该计算在内的�?br>

Output
�Ҏ��l�数据输��Z��行，表示该学生的GPA�Q�保留两位小数。如果GPA不存在，输出-1�?br>

Sample Input
3
3 92
5 80
2 60

Sample Output
2.90

毫无疑问, �q�是一个水�? 但是我悲剧的�q�是WA �?��?, 注意到题目的输出要求里有一句话 : 如果 GPA不存�?输出 -1 .

什么时候不存在? 0 ! 只要把这�? 处理好就行了. YM .

代码如下 :

//MiYu原创, 转帖��h��?: 转蝲�?______________白白の屋

#include <iostream>
#include <iomanip>
using namespace std;
int hash[11] = { 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 4 };
int main ()
{
    int T;
    while ( cin >> T )
    {
            double total = 0 , pnt = 0, scr = 0 , mlt = 0;
            while ( T -- )
            {
                    cin >> pnt >> scr;
                    if ( scr == -1 )
                    {
                         continue;
                    }
                    total += pnt;
                    mlt += pnt * hash[ (int)scr / 10 ];
            }
            if ( total < 1e-5 )
            {
                 cout << -1 << endl;
                 continue;
            }
            cout << setprecision (2) << setiosflags ( ios::fixed ) << mlt * 1.0 / total << endl;
    }
    return 0;
}

MiYu 2010-08-08 20:27 发表评论

HDOJ HDU 2061 Treasure the new start, freshmen! ACM 2061 IN HDU

MiYu — Sat, 07 Aug 2010 14:31:00 GMT

MiYu原创, 转帖��h��?: 转蝲�?______________白白の屋

题目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=2061
题目描述:

Problem Description
background:
A new semester comes , and the HDU also meets its 50th birthday. No matter what's your major, the only thing I want to tell you is:"Treasure the college life and seize the time." Most people thought that the college life should be colorful, less presure.But in actual, the college life is also busy and rough. If you want to master the knowledge learned from the book, a great deal of leisure time should be spend on individual study and practise, especially on the latter one. I think the every one of you should take the learning attitude just as you have in senior school.
"No pain, No Gain", HDU also has scholarship, who can win it? That's mainly rely on the GPA(grade-point average) of the student had got. Now, I gonna tell you the rule, and your task is to program to caculate the GPA.
If there are K(K > 0) courses, the i-th course has the credit Ci, your score Si, then the result GPA is
GPA = (C1 * S1 + C2 * S2 +……+Ci * Si……) / (C1 + C2 + ……+ Ci……) (1 <= i <= K, Ci != 0)
If there is a 0 <= Si < 60, The GPA is always not existed.

Input
The first number N indicate that there are N test cases(N <= 50). In each case, there is a number K (the total courses number), then K lines followed, each line would obey the format: Course-Name (Length <= 30) , Credits(<= 10), Score(<= 100).
Notice: There is no blank in the Course Name. All the Inputs are legal

Output
Output the GPA of each case as discribed above, if the GPA is not existed, ouput:"Sorry!", else just output the GPA value which is rounded to the 2 digits after the decimal point. There is a blank line between two test cases.

Sample Input
2
3
Algorithm 3 97
DataStruct 3 90
softwareProject 4 85
2
Database 4 59
English 4 81

Sample Output
90.10

Sorry!

题目分析:

�q�只是一道简单的数��D��题, 唯一的难度在读题. 直接使用题目�l�出的公式就可以�? 当然如果挂科的话,要输�?"Sorry!"

下面是题目的大致��译:

背景:
新的学期卛_��到来�Q�也正逢杭州电子科技大学50周年校庆。无��Z��的专业是什么，我唯一要告诉那么的是：“珍惜大学的生�z�，利用好时间�?#8221; 很多��为大学的生活回是丰富多采。但事实是，大学生活也同样忙��。如果你要掌握书上的知识�Q�你的空闲时间必��花在自学和实践上，��其是实��c��我认�ؓ�Q�作��Z��位大学生�Q�所有�h都必��ȝ��正自��q��学习态度�?br>“没有不劳而获的事�?#8221;�Q�杭电也讄��了奖学金�Q�你能得到吗�Q�它主要依据是学生获得的GPA(��评�q�_��?。现在，我将告诉你它的规则，你的��d��是�~�程计算GPA�?br>假如有K门课�E�，�W�i门课的学分�ؓCi�Q�你的成�l��ؓ为Si�Q�则GPA为：
GPA = (C1 * S1 + C2 * S2 +……+Ci * Si……) / (C1 + C2 + ……+ Ci……) (1 <= i <= K, Ci != 0)
如果有一门课�E�成�l�在0�?0之间�Q�则GPA��不存在�?/span>

代码如下:

MiYu原创, 转帖��h��?: 转蝲�?______________白白の屋

#include <iostream>
int main ()
{
    int f, T, N;
    double s, c;
    double up, down;
    scanf( "%d", &T );
    while ( T -- )
    {
            scanf ( "%d", &N );
            up = down = f = 0;
            while ( N -- )
            {
                  scanf ( "%*s%lf%lf", &c, &s );
                  if ( f )
                  {
                       continue;
                  }
                  if ( s < 60 && s >= 0 )
                  {
                       f = 1;
                  }
                  up += s * c;
                  down += c;
            }
            printf( f ? "Sorry!\n" : "%.2f\n", up / down );
            if ( T ) putchar ( '\n' );
    }
    return 0;
}

MiYu 2010-08-07 22:31 发表评论

HDOJ HDU 2088 Box of Bricks ACM 2088 IN HDU

MiYu — Sat, 07 Aug 2010 10:08:00 GMT

MiYu原创, 转帖��h��?: 转蝲�?______________白白の屋

题目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=2088
题目描述:

Problem Description
Little Bob likes playing with his box of bricks. He puts the bricks one upon another and builds stacks of different height. “Look, I've built a wall!”, he tells his older sister Alice. “Nah, you should make all stacks the same height. Then you would have a real wall.”, she retorts. After a little consideration, Bob sees that she is right. So he sets out to rearrange the bricks, one by one, such that all stacks are the same height afterwards. But since Bob is lazy he wants to do this with the minimum number of bricks moved. Can you help?

Input
The input consists of several data sets. Each set begins with a line containing the number n of stacks Bob has built. The next line contains n numbers, the heights hi of the n stacks. You may assume 1≤n≤50 and 1≤hi≤100.

The total number of bricks will be divisible by the number of stacks. Thus, it is always possible to rearrange the bricks such that all stacks have the same height.

The input is terminated by a set starting with n = 0. This set should not be processed.

Output
For each set, print the minimum number of bricks that have to be moved in order to make all the stacks the same height.
Output a blank line between each set.

Sample Input
6
5 2 4 1 7 5
0

Sample Output
5

题目分析:
哈哈 ,又是一个水�? 貌似�q�段旉��一直在水题. 刚开始看的时候还以�ؓ�?DP , 看懂题目后知道这只是一个简单的��学数学�?呵呵. 如果求移动步数还复杂�? 但题目要求的�?最��搬动次�? 那么只要求出�q�_��? 大于�q�_��值的墙就是要�U�d��的墙 ,累加��可以了.

代码如下:

MiYu原创, 转帖��h��?: 转蝲�?______________白白の屋

#include <iostream>
#include <string>
using namespace std;
int main ()
{
    int N;
    int f = 0;
    while ( cin >> N, N )
    {
            if ( f )
            {
                 cout << endl;
            }
            f = 1;
            int num[N+1];
            num[N] = 0;
            for ( int i = 0; i != N; ++ i )
            {
                  cin >> num[i];
                  num[N] += num[i];
            }
            num[N] /= N;
            int nCount = 0;
            for ( int i = 0; i != N; ++ i )
            {
                  if ( num[i] > num[N] )
                  {
                       nCount += num[i] - num[N];
                  }
            }
            cout << nCount << endl;
    }
    return 0;
}

MiYu 2010-08-07 18:08 发表评论

HDOJ HDU 2093 考试排名 ACM 2093 IN HDU

MiYu — Sat, 07 Aug 2010 09:33:00 GMT

MiYu原创, 转帖��h��?: 转蝲�?______________白白の屋

题目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=2093
题目描述:

Problem Description
C++�~�程考试使用的实时提交系�l�，��h��x��获得成�W排名的特炏V��它的功能是怎么实现的呢�Q?br>我们做好了题目的解答�Q�提交之后，要么“AC”�Q�要么错误，不管怎样错法�Q��L��l�你��C��一�W�，表明你曾�l�有�q�一�ơ错误提交，因而当你一旦提交该�?#8220;AC”后，��p��与你��一��帐了，��d��该题错误提交了几回。虽然你在题��C��Q�大步地跃上了一个台�Ӟ��但是在耗时上要摊上你共花去的时间。特别是�Q�曾�l�有�q�的错误提交�Q�每�ơ都要摊上一定的单位旉��分。这样一来，你在做出的题��C��Q�可能领先别人很多，但是�Q�在做出同样题数的�h��中�Q�你可能会在耗时上处于排名的劣势�?br>例如�Q�某�ơ考试一�?题（A�Q�B�Q�C�Q�D�Q�E�Q�F�Q�G�Q�H�Q�，每个人做的题都在对应的题号下有个数量标记�Q�负数表�C��学生在该题上有过的错误提交次敎ͼ�但到现在�q�没有AC�Q�正数表�C�AC所耗的旉��Q�如果正数a跟上一�Ҏ��P��里面有个整数b�Q�那��p��C��学生提交该题AC了，耗去了时间a�Q�同�Ӟ��曄��错误提交了b�ơ，因此对于下述输入数据�Q?br>

若每�ơ错误提交的�|�分�?0分，则其排名从高��C��应该是这��L��Q?br>Josephus 5 376
John 4 284
Alice 4 352
Smith 3 167
Bob 2 325
Bush 0 0

Input
输入数据的第一行是考试题数n�Q?/span>1≤n≤12�Q�以及单位罚分数m�Q?/span>10≤m≤20�Q�，每行数据描述一个学生的用户名（不多�?0个字�W�的字串�Q�以及对所有n道题的答题现�Ӟ��其描�q�采用问题描�q�C��的数量标记的格式�Q�见上面的表��|��提交�ơ数��L��于100�Q�AC所耗时间��L��于1000�?br>

Output
��这些学生的考试现状�Q�输��Z��个实时排名。实时排名显然先按AC题数的多��排�Q�多的在前，再按旉��分的多少排，��的在前�Q�如果凑巧前两者都相等�Q�则按名字的字典序排�Q�小的在前。每个学生占一行，输出名字�Q?0个字�W�宽�Q�，做出的题敎ͼ�2个字�W�宽�Q�右寚w��Q�和旉��分（4个字�W�宽�Q�右寚w��Q�。名字、题数和旉��分相互之间有一个空根{�?br>

Sample Input
8 20
Smith      -1    -16    8    0    0    120    39    0
John      116    -2    11    0    0    82    55(1)    0
Josephus  72(3)    126    10    -3    0    47    21(2)    -2
Bush      0    -1    -8    0    0    0    0    0
Alice      -2    67(2)    13    -1    0    133    79(1)    -1
Bob      0    0    57(5)    0    0    168    -7    0

Sample Output
Josephus    5  376
John        4  284
Alice       4  352
Smith       3  167
Bob         2  325
Bush        0    0

题目分析 :
�U��a考验�~�程能力的题, ��׃��l�自己队里的代码牛�h解决�?, 主要是��?sscanf () 用于判断有没有括�?

代码如下:

#include <cstdio>
#include <cstdlib>
#include <cstring>
#define    MAX 1000

typedef struct
{
    char    name[11];
    int    sum;
    int    total;
}credit;

int cmp(const void *a, const void *b)
{
    if ((*(credit *)a).total != (*(credit *)b).total)
        return (*(credit *)a).total - (*(credit *)b).total;
    else if ((*(credit *)b).sum != (*(credit *)a).sum)
        return (*(credit *)b).sum - (*(credit *)a).sum;
    else
        return strcmp((*(credit *)b).name, (*(credit *)a).name);
}

int main()
{
    int    res, t, f;
    int    i, n, m, j;
    char    style[10];
    credit    c[MAX] = {0};
    scanf ( "%d%d", &n, &m );
    for ( i = 0; scanf("%s", c[i].name) != EOF; i++ )
    {
          for ( j = 0; j < n; j++ )
          {
                scanf ( "%s", style );
                res = sscanf(style, "%d(%d)", &t, &f );
                if ( res == 2 )
                {
                    c[i].total++;
                    c[i].sum += t + f * m;
                }
                else if (res == 1 && t > 0)
                {
                    c[i].total++;
                    c[i].sum += t;
                }
          }
    }

    qsort(c, i, sizeof(credit), cmp);
    for ( i--; i >= 0; i-- )
    {
         printf("%-10s %2d %4d\n", c[i].name, c[i].total, c[i].sum);
    }
    return 0;
}

MiYu 2010-08-07 17:33 发表评论

HDOJ HDU 2086 A1 = ? ACM 2086 IN HDU

MiYu — Fri, 06 Aug 2010 08:52:00 GMT

MiYu原创, 转帖��h��?: 转蝲�?______________白白の屋

题目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=2086
题目描述:

Problem Description
有如下方�E�：Ai = (Ai-1 + Ai+1)/2 - Ci (i = 1, 2, 3, . n).
若给出A0, An+1, �?nbsp;C1, C2, ..Cn.
��L��E�计��A1 = ?

Input
输入包括多个��试实例�?br>对于每个实例�Q�首先是一个正整数n,(n <= 3000); 然后�?个数a0, an+1.接下来的n行每行有一个数ci(i = 1, .n);输入以文件结束符�l�束�?br>

Output
对于每个��试实例�Q�用一行输出所求得的a1(保留2位小�?.

Sample Input
1
50.00
25.00
10.00
2
50.00
25.00
10.00
20.00

Sample Output
27.50
15.00

�q�道题的关键在于公式的推�? 公式推出来问题就解决�?

Ai = (Ai-1 + Ai+1)/2 - Ci (i = 1, 2, 3, . n). 用A5做示范，��p��猜测出公式来。证明方法可以采用数学归�U�x��。这里证明从略�?br>A5 = (A4 + A6) / 2 - C5
A4 = (A3 + A5) / 2 - C4 = A3 / 2 + A4 / 4 + A6 / 4 - C5 / 2 - C4
=>A4 = 2A3/3 + A6/3 - 2C5/3 - 4C4/3

A3 = (A2 + A4) / 2 - C3 = A2 / 2 + A3 / 3 + A6 / 6 - C5 / 3 - 2C4 / 3 - C3
=>A3 = 3A2/4 + A6/4 - C5/2 - C4 - 3C3/2

A2 = (A1 + A3) / 2 - C2 = A1 / 2 + 3A2 / 8 + A6 / 8 - C5 / 4 - C4 / 2 - 3C3 / 4 - C2
=>A2 = 4A1/5 + A6/5 - 2C5/5 - 4C4/5 - 6C3/5 - 8C2/5

A1 = (A0 + A2) / 2 - C1 = A0 / 2 + 2A1 / 5 + A6 / 10 - C5 / 5 - 2C4 / 5 - 3C3 / 5 - 4C2 / 5 - C1
=>A1 = 5A0/6 + A6/6 - C5/3 - 2C4/3 - C3 - 4C2/3 - 5C1/3

��到�q�里�Q�我想你已经�ȝ��出公式了:
A1 = (n * A0 + An+1 - 2 * Cn - 4 * Cn-1 - - 2 * i * Cn-i+1 - 2 * n * C1) / (n + 1)

代码如下 :

MiYu原创, 转帖��h��?: 转蝲�?______________白白の屋

#include <stdio.h>
int main ()
{
    int n, i;
    double a0, an1, a1;
    double c[3000];
    while ( scanf ( "%d", &n) != EOF )
    {
        scanf ( "%lf%lf", &a0, &an1 );
        for ( i = 0; i < n; i++ )
        {
             scanf("%lf", c + i);
        }
        a1 = n * a0 + an1;
        for ( i = 1; i <= n; i++ )
        {
              a1 -= 2 * i * c[n - i];
        }
        printf ( "%.2f\n", a1 / ( n + 1 ) );
    }
    return 0;
}

MiYu 2010-08-06 16:52 发表评论