hdu1538 A Puzzle for Pirates---海盗分金---博弈

A Puzzle for PiratesTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 141    Accepted Submission(s): 32


Problem Description

Your task is to predict how many gold pieces a given pirate will get. InputOutputSample InputSample OutputAuthorSourceRecommend#include<iostream>#include<cstdio>#include<ctime>#include<cstring>#include<cmath>#include<algorithm>#include<cstdlib>#include<vector>#define C 240#define TIME 10#define inf 1<<25#define LL long longusing namespace std;//保存2的幂int fac[15]={2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768};void slove(int n,int m,int p){ //金币够贿赂的情况 if(n<=2*m){ //不是决策者，而且奇偶性相同，都能被贿赂 if(n!=p&&(n%2==p%2)) printf("1\n"); //剩下的都是决策者拥有 else if(n==p) printf("%d\n",m-(n-1)/2); else //其余人分不到金币，他们的决策不影响全局 printf("0\n"); return ; } //这时候的不同在于决策者不能拿金币 else if(n==2*m+1){ if(p<2*m&&p&1) printf("1\n"); else printf("0\n"); return ; } int t=n-2*m,i; //这是刚好保命的情况，对于决策者来说，肯定没有金币 //对于其它人来说，要么肯定没有金币，要么可能没有金币，不确定性 for( i=0;i<14;i++){ if(t==fac[i]&&p==t)//这应该加上p==t { printf("0\n"); return; } } // int i; for( i=1;i<14;i++) if(t<fac[i]) { //决策者必死 if(p>2*m+fac[i-1]&&p<2*m+fac[i]) printf("Thrown\n"); else printf("0\n"); return ; }}int main(){ int t,n,m,p; scanf("%d",&t); while(t--){ scanf("%d%d%d",&n,&m,&p); slove(n,m,p); } return 0;}