NYOJ364田忌赛马

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题目链接：http://acm.nyist.net/JudgeOnline/problem.php?pid=364

思路：1.当田忌的快马比齐王的快马块时，赢一场

2.当田忌的慢马比齐王的慢马快时，赢一场

3.当1和2都不满足时，用田忌的慢马去和齐王的慢马去比。

代码：

#include <cstdio>#include <iostream>#include <algorithm>using namespace std;int tian[1005];int king[1005];int main(){    int n;    while(~scanf("%d",&n))    {        for(int i = 0;i < n;++i)            scanf("%d",&tian[i]);        for(int i = 0;i < n;++i)            scanf("%d",&king[i]);        sort(tian,tian + n);        sort(king,king + n);        int win = 0;        int lose = 0;        int tl = 0,tr = n - 1;        int kl = 0,kr = n - 1;        while(tl <= tr && kl <= kr)        {            if(tian[tr] > king[kr])            {                win++;                tr--;                kr--;            }            else if(tian[tl] > king[kl])            {                 win++;                 tl++;                 kl++;            }            else            {                if(tian[tl] == king[kr])                {                    tl++;                    kr--;                }                else if(tian[tl] < king[kr])                {                    lose++;                    tl++;                    kr--;                }            }        }        printf("%d\n",(win - lose) * 200);    }    return 0;}

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