hdu2176 取(m堆)石子游戏（nim博弈）

取(m堆)石子游戏

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2713    Accepted Submission(s): 1617

Problem Description

m堆石子,两人轮流取.只能在1堆中取.取完者胜.先取者负输出No.先取者胜输出Yes,然后输出怎样取子.例如5堆 5,7,8,9,10先取者胜,先取者第1次取时可以从有8个的那一堆取走7个剩下1个,也可以从有9个的中那一堆取走9个剩下0个,也可以从有10个的中那一堆取走7个剩下3个.

 

Input

输入有多组.每组第1行是m,m<=200000. 后面m个非零正整数.m=0退出.

 

Output

先取者负输出No.先取者胜输出Yes,然后输出先取者第1次取子的所有方法.如果从有a个石子的堆中取若干个后剩下b个后会胜就输出a b.参看Sample Output.

 

Sample Input

245 4533 6 955 7 8 9 100

 

Sample Output

NoYes9 5Yes8 19 010 3

 

Author

Zhousc

 

Source

ECJTU 2008 Summer Contest

 

 

题意：中文题.....不说了....

分析：这不和hdu1850差不多嘛；稍微改一下不就行了

#include <iostream>#include <cstdio>#include <cstring>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <cmath>#include <algorithm>using namespace std;const double eps = 1e-6;const double pi = acos(-1.0);const int INF = 1e9;const int MOD = 1e9+7;#define ll long long#define CL(a,b) memset(a,b,sizeof(a))#define lson (i<<1)#define rson ((i<<1)|1)#define N 50010int gcd(int a,int b){    return b?gcd(b,a%b):a;}int n,a[200000+10];int main(){    while(scanf("%d",&n),n)    {        int sum=0,ans=0;        for(int i=0; i<n; i++)        {            scanf("%d",&a[i]);            sum ^= a[i];        }        if(sum==0) cout<<"No"<<endl;        else        {            cout<<"Yes"<<endl;            for(int i=0; i<n; i++)            {                if((sum^a[i]) <= a[i])                     cout<<a[i]<<" "<<(sum^a[i])<<endl;            }        }    }    return 0;}