Codeforces - 662A 思路巧妙的异或

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<bitset>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define rrep(i,j,k) for(register int i=j;i>=k;i--)
#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])
#define iin(a) scanf("%d",&a)
#define lin(a) scanf("%lld",&a)
#define din(a) scanf("%lf",&a)
#define s0(a) scanf("%s",a)
#define s1(a) scanf("%s",a+1)
#define print(a) printf("%lld",(ll)a)
#define enter putchar(‘\n‘)
#define blank putchar(‘ ‘)
#define println(a) printf("%lld\n",(ll)a)
#define IOS ios::sync_with_stdio(0)
using namespace std;
const int MAXN = 5e5+11;
const double EPS = 1e-7;
typedef long long ll;
typedef unsigned long long ull;
const ll MOD = 10086; 
unsigned int SEED = 17;
const ll INF = 1ll<<60;
ll read(){
    ll x=0,f=1;register char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
ll b[66];
int cal(int n,ll a[]){
    memset(b,0,sizeof b);
    int cnt=0;
    rep(i,1,n){
        rrep(j,62,0){
            if(a[i]>>j&1){
                if(b[j]) a[i]^=b[j];
                else{
                    b[j]=a[i];
                    rrep(k,j-1,0) if(b[k]&&(b[j]>>k&1))b[j]^=b[k];
                    rep(k,j+1,62) if(b[k]>>j&1) b[k]^=b[j];
                    break;
                }
            }
        }
    }
    rep(i,0,62) if(b[i]) cnt++;
    return cnt;
}
ll A[MAXN],B[MAXN],C[MAXN],n;
int main(){
    while(cin>>n){
        rep(i,1,n){
            A[i]=read();
            B[i]=read();
        }
        ll S=0;
        rep(i,1,n) S^=A[i],C[i]=A[i]^B[i];
        int cnt=cal(n,C);
        rep(i,0,62) if(S>>i&1) S^=b[i];
        //注意如果i位没有别瞎异或,相当于构造时的插入但不更新的操作
        if(S){//不在线性基中
            printf("1/1\n");
        }else{
            ll ans=1ll<<cnt; //线性基的所有可能
            printf("%lld/%lld\n",ans-1,ans); //把唯一存在的异或为S的剔除便是胜率
        }
    }
    return 0;
}