【Codeforces 662A】Gambling Nim

题目传送门：http://codeforces.com/problemset/problem/662/A
官方题解：http://codeforces.com/blog/entry/44408

说人话就是亦或和为0的概率是多少
考虑把ai全部亦或起来为sum的话
那就是求{ai^bi}有多少个子集亦或和为sum
这样的话，岂不就是线性基的拿手好戏了？

#include<bits/stdc++.h>
#define LL long long
using namespace std;

const int N = 500000+9;
const int SGZ = 61+9;

int n,cnt;
LL a[N],b[N],bas[SGZ],sum; 

inline LL read(){
	char c=getchar(); LL ret=0,f=1;
	while (c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}
	while (c<='9'&&c>='0') {ret=ret*10+c-'0';c=getchar();}
	return ret*f;
}

inline bool insert(LL w) {
	for (int i=61;~i;i--) if (w&(1LL<<i)) {
		if (bas[i]) {
			w ^= bas[i];
		} else {
			bas[i] = w;
			return true;
		}
	}	
	return false;
}

inline LL pow(LL w, int t) {
	LL ret = 1;
	while (t) {
		if (t & 1) ret *= w;
		w *= w; t >>= 1;
	}
	return ret;
}

int main(){
	n = read();
	for (int i=1;i<=n;i++) {
		a[i] = read();
		b[i] = read();
		sum ^= a[i];
		cnt += insert(a[i] ^ b[i]);
	}
	if (!cnt && !sum) {
		puts("0/1");
	} else if (insert(sum)) {
		puts("1/1");
	} else {
		printf("%I64d/%I64d\n",pow(2LL,cnt)-1,pow(2LL,cnt));
	}
	return 0;
}