hdu 1729 Stone Game（SG函数）

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本文探讨了两人参与的游戏——石子博弈的策略。玩家轮流选择盒子并放入一定数量的石子，规则限制了每一步的操作，最终无法再放置石子的玩家将输掉比赛。通过输入盒子的数量和初始石子数，文章分析了第一玩家是否能够赢得游戏，基于最优策略进行判断。

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Stone Game

Problem Description

This game is a two-player game and is played as follows:

1. There are n boxes; each box has its size. The box can hold up to s stones if the size is s.
2. At the beginning of the game, there are some stones in these boxes.
3. The players take turns choosing a box and put a number of stones into the box. The number mustn’t be great than the square of the number of stones before the player adds the stones. For example, the player can add 1 to 9 stones if there are 3 stones in the box. Of course, the total number of stones mustn’t be great than the size of the box.
4.Who can’t add stones any more will loss the game.

Give an Initial state of the game. You are supposed to find whether the first player will win the game if both of the players make the best strategy.

Input

The input file contains several test cases.
Each test case begins with an integer N, 0 < N ≤ 50, the number of the boxes.
In the next N line there are two integer si, ci (0 ≤ ci ≤ si ≤ 1,000,000) on each line, as the size of the box is si and there are ci stones in the box.
N = 0 indicates the end of input and should not be processed.

Output

For each test case, output the number of the case on the first line, then output “Yes” (without quotes) on the next line if the first player can win the game, otherwise output “No”.

Sample Input

Sample Output

  
   Case 1:
Yes
Case 2:
No

Source

“网新恩普杯”杭州电子科技大学程序设计邀请赛

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#include<cstdio>
#include<cmath>
using namespace std;
int SG(int s,int c)
{
    int t=sqrt(s);
    while(t*t+t>=s) t--;//t*t+t<s (1+t)^2+(1+t)>=s
    if(c>t) return s-c;//必胜
    else return SG(t,c);//(t,c) 与(s,c)情况等价
}
int main()
{
    int kase=1,ans,c,s,n;
    while(scanf("%d",&n)&&n)
    {
        ans=0;
        while(n--)
        {
            scanf("%d%d",&s,&c);
            ans^=SG(s,c);
        }
        printf("Case %d:\n",kase++);
        if(ans)
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}