---
title: Gena vs Petya
tags:
  - Dynamic Programming
  - Game Theory
id: "2670"
categories:
  - [OI, Common Skill]
  - [OI, Number Theory]
date: 2018-04-30 03:15:41
---

## 题目描述

Gena and Petya love playing the following game with each other. There are $n$ piles of stones, the $i$-th pile contains $a_i$ stones. The players move in turns, Gena moves first. A player moves by choosing any non-empty pile and taking an arbitrary positive number of stones from it. If the move is impossible (that is, all piles are empty), then the game finishes and the current player is considered a loser.

Gena and Petya are the world famous experts in unusual games. We will assume that they play optimally.

Recently Petya started to notice that Gena wins too often. Petya decided that the problem is the unjust rules as Gena always gets to move first! To even their chances, Petya decided to cheat and take and hide some stones before the game begins. Since Petya does not want Gena to suspect anything, he will take the same number of stones $x$ from each pile. This number $x$ can be an arbitrary non-negative integer, strictly less that the minimum of $a_i$ values.

Your task is to find the number of distinct numbers $x$ such that Petya will win the game.

## 题意概述

给定$n$个正整数$a_i$，求有多少个非负整数$x$，满足$x$小于所有给定的数，且所有给定的数减去$x$之后的异或和为$0$。

数据范围：$1 \le n \le 2 \times 10^5, \; 1 \le a_i \le 10^{18}$。

## 算法分析

Nim 游戏后手胜利的条件是所有数的异或和为$0$。

枚举$x$显然不可行。可以尝试从低位到高位依次确定。

考虑低位对高位的影响，易知只有当低位退位时才会改变高位。而对于每一位，若此位上有偶数个$0$，则此位可以填$1$，若有偶数个$1$则可以填$0$（这样才能使得异或和为$0$）。

令$f_{i, j}$表示从低到高处理到第$i$位时，有$j$个数字在第$(i+1)$位退位的方案数。根据退位的性质，这$j$个数字一定是所有给定的数按后$i$位从小到大排序后的前$j$个。可以在枚举$i$的同时基数排序，枚举$j$的同时维护第$i$位上$0$和$1$的个数（会受退位影响），若满足条件则转移（注意此位填$1$时还要考虑此位$0$的退位）。

最后需要特判一下$x$不能等于所有给定的数中最小的数。

## 代码

```cpp
#include <algorithm>
#include <cstdio>
#include <cstring>

#define int long long

static int const N = 200005;
static int const M = 65;
int a[N], cnt[M], f[M][N], s[2][N];

signed main() {
  int n;
  scanf("%lld", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%lld", a + i);
    for (int j = 0; j < M; ++j)
      cnt[j] += a[i] >> j & 1;
  }
  f[0][0] = 1;
  for (int i = 0; i < M - 1; ++i) {
    int c0 = n - cnt[i], c1 = cnt[i], c = 0;
    for (int j = 0; j <= n; ++j) {
      if (j)
        if (a[j - 1] >> i & 1)
          ++c0, --c1;
        else
          --c0, ++c1, ++c;
      if (!(c0 & 1))
        f[i + 1][c + c0] += f[i][j];
      if (!(c1 & 1))
        f[i + 1][c] += f[i][j];
    }
    *s[0] = *s[1] = 0;
    for (int j = 0; j < n; ++j)
      s[a[j] >> i & 1][++*s[a[j] >> i & 1]] = a[j];
    for (int j = 1; j <= *s[0]; ++j)
      a[j - 1] = s[0][j];
    for (int j = 1; j <= *s[1]; ++j)
      a[*s[0] + j - 1] = s[1][j];
  }
  int sum = 0;
  for (int i = 1; i < n; ++i)
    sum ^= a[i] - a[0];
  printf("%lld\n", f[M - 1][0] - (sum == 0));
  return 0;
}
```
