hdu1729 Stone Game

Stone Game

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2263 Accepted Submission(s): 645

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

3 2 0 3 3 6 2 2 6 3 6 3 0

Sample Output

Case 1: Yes Case 2: No

可以转化成取石子游戏,这里,我们可以求出每一堆的sg值 ,就可以看成取石子了,怎么求sg值呢,我们首先知道,对于n,k,一定存在 一个p使得p+p*p=n;这样,我们就可以判定,如果k==p一定是负,因为不能一步到达,k>p,刚可以一步达,返回n-k;k

#include ()
#include 
#include 
using namespace std;
int getsg(int n,int k){
    int p=sqrt(n);
    while(p+p*p>=n)
    p--;
    if(p