扑克游戏

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题目不难,但是仍然要想清除;

易得赢得总数等于输的总数;

例如输入的为B、C、D

其次最大赢家数和最小赢家数一定为一组,因此,排好序,未知量只能是A或者E;即A和D一组或者B和E一组;然后分情况讨论满不满足即可;

// 扑克游戏.cpp : 定义控制台应用程序的入口点。////#include "stdafx.h"#include<iostream>#include<algorithm>#include<cstdio>using namespace std;const int maxn = 10005;struct node{int x, y;bool operator<( const node &a )const{return x < a.x;}}edge[ maxn ];//int _tmain(int argc, _TCHAR* argv[])int main(){while( cin >> edge[ 0 ].x >> edge[ 0 ].y >> edge[ 1 ].x >> edge[ 1 ].y >> edge[ 2 ].x >> edge[ 2 ].y ){if( !edge[ 0 ].x && !edge[ 0 ].y && !edge[ 1 ].x && !edge[ 1 ].y && !edge[ 2 ].x && !edge[ 2 ].y )break;sort( edge, edge + 2 );int y = edge[ 1 ].x + edge[ 2 ].x - edge[ 0 ].y;int x = edge[ 1 ].y + edge[ 2 ].y - edge[ 0 ].x;if( edge[ 0 ].x + edge[ 1 ].x + edge[ 2 ].x + x != edge[ 0 ].y + edge[ 1 ].y + edge[ 2 ].y + y ){x = edge[ 0 ].y + edge[ 1 ].y - edge[ 2 ].x;y = edge[ 0 ].x + edge[ 1 ].x - edge[ 2 ].y;}cout << x << "-" << y << endl;}return 0;}