A Simple Game

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/65535 K (Java/Others)

Total Submission(s): 75 Accepted Submission(s): 52

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 M_n 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 L_n 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 M_i and L_i (20 ≥ M_i > 0, 20 ≥ L_i > 0).

Output

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

Sample Input

Sample Output

Yes
No

Author

Agreal@TJU

Source

HDU 2007 Programming Contest - Final

Recommend

lcy

/*
题意：有m堆石子，每堆石子有Mi个，每次最多取Li个，谁取到最后一个谁输

初步思路：对于一堆来说是八神博弈，对于整体来说是尼姆博弈
*/
#include<bits/stdc++.h>
using namespace std;
int t,n,m,l;
int res=0;
int main(){
    // freopen("in.txt","r",stdin);
    scanf("%d",&t);
    while(t--){
        res=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%d%d",&m,&l);
            res^=(m%(l+1));
        }
        printf(res?"No
":"Yes
");
    }
    return 0;
}