#include<stdio.h>
#include<string.h>
int main()
{
    int n,count,temp,count1,sum;
    int i,j,a[1010],b[1010];
    while(scanf("%d",&n)&&(n!=0))
    {
        sum=count=count1=0;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
         for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
           for(i=0;i<n;i++)
        {
            scanf("%d",&b[i]);
        }
        for(i=0;i<n-1;i++)
        for(j=i+1;j<n;j++)
        {
          if(a[i]>a[j])
           {
            temp=a[i];
            a[i]=a[j];
            a[j]=temp;
          }
          if(b[i]>b[j])
           {
            temp=b[i];
            b[i]=b[j];
            b[j]=temp;
          }
        }

        j=0;
        for(i=0;i<n;i++)
        for(;j<n;)
        {
            if(a[i]>b[j])
            {

                count++;
                j++;
                    break;

            }
            if(a[i]==b[j])
            {
                if((a[i+1]<=b[j+1])&&(a[i+1]>b[j]))
                {
                    count++;
                    i=i+1;
                    j=j+1;
                    break;
                }
                count1++;
                j++;
                break;
            }
            if(count1!=0)
              {count++;
               count1--;}
             break;
        }
        sum=count-(n-count-count1);
        printf("%d\n",sum*200);
    }
  return 0;
}











Tian Ji -- The Horse Racing
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 2135 Accepted Submission(s): 565  

 
Problem Description
Here is a famous story in Chinese history.

"That was about 2300 years ago. General Tian Ji was a high official in the country Qi. He likes to play horse racing with the king and others."

"Both of Tian and the king have three horses in different classes, namely, regular, plus, and super. The rule is to have three rounds in a match; each of the horses must be used in one round. The winner of a single round takes two hundred silver dollars from the loser."

"Being the most powerful man in the country, the king has so nice horses that in each class his horse is better than Tian's. As a result, each time the king takes six hundred silver dollars from Tian."

"Tian Ji was not happy about that, until he met Sun Bin, one of the most famous generals in Chinese history. Using a little trick due to Sun, Tian Ji brought home two hundred silver dollars and such a grace in the next match."

"It was a rather simple trick. Using his regular class horse race against the super class from the king, they will certainly lose that round. But then his plus beat the king's regular, and his super beat the king's plus. What a simple trick. And how do you think of Tian Ji, the high ranked official in China?"



Were Tian Ji lives in nowadays, he will certainly laugh at himself. Even more, were he sitting in the ACM contest right now, he may discover that the horse racing problem can be simply viewed as finding the maximum matching in a bipartite graph. Draw Tian's horses on one side, and the king's horses on the other. Whenever one of Tian's horses can beat one from the king, we draw an edge between them, meaning we wish to establish this pair. Then, the problem of winning as many rounds as possible is just to find the maximum matching in this graph. If there are ties, the problem becomes more complicated, he needs to assign weights 0, 1, or -1 to all the possible edges, and find a maximum weighted perfect matching...

However, the horse racing problem is a very special case of bipartite matching. The graph is decided by the speed of the horses --- a vertex of higher speed always beat a vertex of lower speed. In this case, the weighted bipartite matching algorithm is a too advanced tool to deal with the problem.

In this problem, you are asked to write a program to solve this special case of matching problem.

 
 
Input
The input consists of up to 50 test cases. Each case starts with a positive integer n (n <= 1000) on the first line, which is the number of horses on each side. The next n integers on the second line are the speeds of Tian’s horses. Then the next n integers on the third line are the speeds of the king’s horses. The input ends with a line that has a single 0 after the last test case.

 
 
Output
For each input case, output a line containing a single number, which is the maximum money Tian Ji will get, in silver dollars.

 
 
Sample Input
3
92 83 71
95 87 74
2
20 20
20 20
2
20 19
22 18
0
 
Sample Output
200
0
0

我给你改了改，有些地方不知道你什么意思就自己写了（因为你没注释…），带//&&&&&&&的地方是我写的，/*....*/的地方是把你的空出来的。


#include<stdio.h>
#include<string.h>
int main()
{
    int n,count,temp,count1,sum;
    int i,j,x=0,d=0,a[1010],b[1010],c[100];            //&&&&&&&&&&多加了一个数组和判断胜负的变量
    while(scanf("%d",&n)&&(n!=0))
    {
        sum=count=count1=0;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
         for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
           for(i=0;i<n;i++)
        {
            scanf("%d",&b[i]);
        }
        for(i=0;i<n-1;i++)
            for(j=i+1;j<n;j++)
            {
              if(a[j]>a[j-1])                    //&&&&&&&&&&&你这里的冒泡排序下标错了，可能你没
               {                                //&&&&&&&&&&&弄出结果有这部分原因
                temp=a[j-1];
                a[j-1]=a[j];
                a[j]=temp;
              }
              if(b[j-1]<b[j])
               {
                temp=b[j-1];
                b[j-1]=b[j];
                b[j]=temp;
              }
            }

        i=j=d=0;                                //&&&&&&&&&&&&&&&&实在不懂你要怎么判断胜负，自己写了个
            while(j!=n)
            {
                if(a[i]<b[j])
                {
                    j++;
                }
                else if(a[i]==b[j])
                {
                    d++;
                    j++;
                }
                else
                {    i++;
                    j++;
                    count++;
                }                                //&&&&&&&&&&&&&&&&&
                /*if(a[i]>b[j])
                {

                    count++;
                    j++;
                    break;

                }
                if(a[i]==b[j])
                {
                    if((a[i+1]<=b[j+1])&&(a[i+1]>b[j]))
                    {
                        count++;
                        i=i+1;
                        j=j+1;
                        break;
                    }
                    count1++;
                    j++;
                    break;
                }
                if(count1!=0)
                  {count++;
                   count1--;}
                 break;*/
            }
        c[x]=count-(n-i-d);                        //&&&&&&&&&&净胜的场数存入c[]中
        x++;
      
    }
    printf("\n");                                //&&&&&&&&&&&我把输出摆在所有输入进行完了之后
    for(i=0;i<x;i++)                            //&&&&&&&&&&&这样比较像题目给的
     printf("%d\n",c[i]*200);
  return 0;
}

楼主，对你的程序做了一些修改，你看看：

#include<stdio.h>
#include<string.h>
int main()
{
    int n,count,temp,count1,sum;
    int i,j,a[1010],b[1010];
    while(scanf("%d",&n)&&(n!=0))
    {
        sum=count=count1=0;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
         for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
           for(i=0;i<n;i++)
        {
            scanf("%d",&b[i]);
        }
        for(i=0;i<n-1;i++)
        for(j=i+1;j<n;j++)        //选择排序法
        {
 //         if(a[i]>a[j])    //结果是从小到大排序了
          if(a[i]<a[j])        //要改为从大到小排序
           {
            temp=a[i];
            a[i]=a[j];
            a[j]=temp;
          }
//          if(b[i]>b[j])
             if(b[i]<b[j])        //改为从大到小排序
           {
            temp=b[i];
            b[i]=b[j];
            b[j]=temp;
          }
        }
        j=0;
        for(i=0;i<n;i++){        //a[i]表示田忌的马队
        for(;j<n;j++)                //b[j]表示齐王的马队
        {
            if(a[i]>b[j])        //如果田忌胜齐王
            {
                count++;        //赢数加1
                j++;            //为下一次比较做准备
                    break;        //跳出内循环
            }
            if(a[i]==b[j])        //这里的情况有些复杂
            {
//                if((a[i+1]<=b[j+1])&&(a[i+1]>b[j]))
                if((i+1<n)&&(j+1<n)&&(a[i+1]>b[j+1]))    //如果双方都有下一场，且下一场能胜
                {
//                    count++;
                    count1++;    //这一场为平，平的场数加1
//                    i=i+1;    //跳出内循环后，在外循环中会自动加1，这里就免了
                    j=j+1;
                    break;
                }
 //               count1++;        此处不用计算负的场数
 //               j++;
 //               break;
            }
//            if(count1!=0)
//              {count++;
//               count1--;}
//             break;
        }
    if(j>=n)break;
      }
      sum=count-(n-count-count1);
      printf("%d\n",sum*200);
    }
return 0;
}

（为了让大家都能看懂这个问题，我把这个题目的描述附上）

关于田忌赛马的程序的描述：

田忌与齐王赛马，双方各有n匹马参赛（n<=100），每场比赛赌注为200两黄金，现已知齐王与田忌的每匹马的速度，并且齐王肯定是按马的速度从快到慢出场，现要你写一个程序帮助田忌计算他最好的结果是赢多少两黄金（输用负数表示）。


[ 本帖最后由 netlin 于 2012-7-23 19:58 编辑 ]