Divide by Zero 2017 and Codeforces Round #399 (Div. 1 + Div. 2, combined) E. Game of Stones Nim游戏

E. Game of Stones

链接：

http://codeforces.com/contest/768/problem/E

题解：

状态数为 n(n+1)/2 <=x 的最大x，之后就进行异或和即可。

代码 ：

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int, int> PII;
const ll mod = 1000000007;
// head

int n, a[66];
int main()
{
    int st = 0, an = 0;
    a[0] = 0;
    rep(i, 1, 10) rep(j, 1, i + 1) a[++st] = i;
    scanf("%d", &n);
    while (n--){
        scanf("%d", &st);
        an ^= a[st];
    }
    if (an == 0) printf("YES
");
    else printf("NO
");
    return 0;
}