hdu-1338 game predictions(贪心题)
Posted smallhester
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了hdu-1338 game predictions(贪心题)相关的知识,希望对你有一定的参考价值。
Given your cards received at the beginning, write a program to tell the maximal number of rounds that you may at least win during the whole game.
InputThe input consists of several test cases. The first line of each case contains two integers m (2 <= m <= 20) and n (1 <= n <= 50), representing the number of players and the number of cards each player receives at the beginning of the game, respectively. This followed by a line with n positive integers, representing the pips of cards you received at the beginning. Then a blank line follows to separate the cases.
The input is terminated by a line with two zeros.
OutputFor each test case, output a line consisting of the test case number followed by the number of rounds you will at least win during the game.
Sample Input
2 5 1 7 2 10 9 6 11 62 63 54 66 65 61 57 56 50 53 48 0 0
Sample Output
Case 1: 2 Case 2: 4
题意:模拟打牌,m个人,每人n张牌。点数1-n*m,出得最大的那个人赢一局,问你最多赢几局
题解:贪心,我出大的,对面出最小的,我出小的,对面最小但比我大的
#include<iostream> #include<algorithm> #include<cstring> #include<sstream> #include<cmath> #include<cstdlib> #include<queue> using namespace std; #define PI 3.14159265358979323846264338327950 bool a[1050]; int main() { int m=0,n=0,count=0,t; while(scanf("%d %d",&m,&n) && (m||n)) { count++; int i=0; int win=0,big=0; memset(a,0,sizeof(a)); for(i=0;i<n;i++) { scanf("%d",&t); a[t]=1; } for(i=n*m;i>0;i--) { if (a[i]) { if (big==0) ++win; else big--; } else big++; } printf("Case %d: %d ",count,win); } return 0; }
以上是关于hdu-1338 game predictions(贪心题)的主要内容,如果未能解决你的问题,请参考以下文章