UESTC - 878

状态的枚举还需多多练习啊

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#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 = 5e4+11;
const int oo = 0x3f3f3f3f;
const double eps = 1e-7;
typedef long long ll;
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 dp[18][130][130],a[18],n;
int main(){
    while(cin>>n){
        rep(i,1,n) a[i]=read();
        memset(dp,0,sizeof dp);
        dp[0][0][0]=1;
        rep(i,1,n){
            rep(j,0,128){
                rep(k,0,128){
                    dp[i][j][k]+=dp[i-1][j][k];
                    dp[i][j^a[i]][k]+=dp[i-1][j][k];//zuo na
                    dp[i][j][k^a[i]]+=dp[i-1][j][k];//you na
                }
            }
        }
        ll ans=0;
        rep(i,0,128){
            rep(j,i,128){
                ans+=dp[n][i][j];
            }
        }
        println(ans);
    }
    return 0;
}