hdu 1851 A Simple Game （Nim博弈）

A Simple Game

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 1401 Accepted Submission(s): 887

Problem Description
Agrael likes play a simple game with his friend Animal during the classes. In this Game there are n piles of stones numbered from 1 to n, the 1st pile has M1 stones, the 2nd pile has M2 stones, … and the n-th pile contain Mn stones. Agrael and Animal take turns to move and in each move each of the players can take at most L1 stones from the 1st pile or take at most L2 stones from the 2nd pile or … or take Ln stones from the n-th pile. The player who takes the last stone wins.

After Agrael and Animal have played the game for months, the teacher finally got angry and decided to punish them. But when he knows the rule of the game, he is so interested in this game that he asks Agrael to play the game with him and if Agrael wins, he won’t be punished, can Agrael win the game if the teacher and Agrael both take the best move in their turn?

The teacher always moves first(-_-), and in each turn a player must takes at least 1 stones and they can’t take stones from more than one piles.

Input
The first line contains the number of test cases. Each test cases begin with the number n (n ≤ 10), represent there are n piles. Then there are n lines follows, the i-th line contains two numbers Mi and Li (20 ≥ Mi > 0, 20 ≥ Li > 0).

Output
Your program output one line per case, if Agrael can win the game print “Yes”, else print “No”.

Sample Input

2
1
5 4
2
1 1
2 2

Sample Output

Yes
No

题意：给你 n 堆石子，然后给你每堆石子的个数和这堆石子每次最多取得个数 m ，谁最后取完谁赢。

思路：通过 sg 函数可以看出规律 sg[n]=n%(m+1) .

ac代码：

/* ***********************************************
Author       : AnICoo1
Created Time : 2016-08-07-17.35 Sunday
File Name    : D:\MyCode\2016-8\2016-8-6-game.cpp
LANGUAGE     : C++
Copyright  2016 clh All Rights Reserved
************************************************ */
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x,y) memset(x,(y),sizeof(x))
#define PI acos(-1)
#define eps 1e-8
using namespace std;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}
//head
int main()
{
    int t;cin>>t;
    while(t--)
    {
        int n;cin>>n;int ans=0;
        for(int i=0;iint a,b;cin>>a>>b;
            ans^=a%(b+1);
        }
        if(ans) printf("No\n");
        else printf("Yes\n");
    }
    return 0;
}