codeforces 768E Game of Stones

E. Game of Stones
time limit per test3 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Sam has been teaching Jon the Game of Stones to sharpen his mind and help him devise a strategy to fight the white walkers. The rules of this game are quite simple:

The game starts with n piles of stones indexed from 1 to n. The i-th pile contains si stones.
The players make their moves alternatively. A move is considered as removal of some number of stones from a pile. Removal of 0 stones does not count as a move.
The player who is unable to make a move loses.
Now Jon believes that he is ready for battle, but Sam does not think so. To prove his argument, Sam suggested that they play a modified version of the game.

In this modified version, no move can be made more than once on a pile. For example, if 4 stones are removed from a pile, 4 stones cannot be removed from that pile again.

Sam sets up the game and makes the first move. Jon believes that Sam is just trying to prevent him from going to battle. Jon wants to know if he can win if both play optimally.

Input
First line consists of a single integer n (1 ≤ n ≤ 106) — the number of piles.

Each of next n lines contains an integer si (1 ≤ si ≤ 60) — the number of stones in i-th pile.

Output
Print a single line containing “YES” (without quotes) if Jon wins, otherwise print “NO” (without quotes)

Examples
input
1
5
output
NO
input
2
1
2
output
YES
Note
In the first case, Sam removes all the stones and Jon loses.

In second case, the following moves are possible by Sam:

In each of these cases, last move can be made by Jon to win the game as follows:

解题思路:
【题意】
有一堆石头每次从中取任意个，但不能取之前取过的数字，问后手的胜负。

【类型】
博弈论

【分析】
首先，这道题的解题思想是这样的——
分析改变后的规则，每种数字只有有限次的取出可能，所以先求出它最大取出次数的可能。
每次都取能取的最小的就能求出最大取出数

for(int i=1;i<60;i++){b[i]=b[i-1]+i;}

通用可以求出式子

k = ((int)sqrt(8*n+1)-1)/2;

题目链接→Codeforces 768E Game of Stones

AC代码

#include<algorithm>
#include <iostream>
#include  <ctype.h>
#include  <cstring>
#include  <fstream>
#include   <cstdio>
#include   <vector>
#include   <string>
#include    <cmath>
#include    <stack>
#include    <queue>
#include      <set>
#include      <map>
#define INF (1<<30)
#define PI acos(-1.0)
typedef long long ll;
using namespace std;
const int N = 1e6 + 5;
int b[100];
int main() {
    int n;
    b[0]=0;
    scanf("%d",&n);
    for(int i=1;i<60;i++){
        b[i]=b[i-1]+i;
    }
    int s,sg=0;
    for(int i=0;i<n;i++)
    {
        scanf("%d",&s);
        for(int j=0;j<20;j++)
        {
            if(s<b[j]){
                sg^=(j-1);
                break;
            }
        }
    }
    if(sg)
    {
        printf("NO\n");
    }
    else{
        printf("YES\n");
    }
    return 0;
}