BZOJ4017 小Q的无敌异或(位运算)

题目链接 小Q的无敌异或

好久之前做的这道题了……参照了别人的博客……还是没有全懂。

第一个问题维护个前缀就好了，第二个问题还要用树状数组维护……

#include <bits/stdc++.h>

using namespace std;

#define rep(i,a,b)              for(int i(a); i <= (b); ++i)
#define LL              long long
#define mod            998244353

const int N     =    100000      +       10;
const int A     =    30          +       1;


LL bit[N];
int a[N], c[N];
int n;
int cnt[A];
LL xn[N], sum[N];
int tmp;
LL p[N];
LL ans1, ans2;

inline void update(int x){ for ( ; x <= n; x += ~x & x + 1) c[x] ^= 1;}
inline int query(int x){ int ret = 0; for ( ; x >= 0; x -= ~x & x + 1) ret ^= c[x]; return ret;}

inline int idx(LL x){
    int l = -1, r = n;
    for (; l < r;){
        int mid = l + r + 1 >> 1;
        if (p[mid] <= x) l = mid; else r = mid - 1;
    }
    return l;
}


int main(){

    bit[0] = 1;
    rep(i, 1, 53) bit[i] = bit[i - 1] * 2;
    scanf("%d", &n);
    rep(i, 1, n){
        scanf("%d", a + i);
        xn[i] = xn[i - 1] ^ a[i];
        sum[i] = sum[i - 1] + a[i];
    }

    rep(k, 0, 30){
        cnt[0] = cnt[1] = 0; tmp = 0;
        rep(i, 0, n){
            (tmp += cnt[((xn[i] >> k) & 1) ^ 1]) %= mod;
            ++cnt[(xn[i] >> k) & 1];
        }

        (ans1 += (LL)(bit[k] * tmp) % mod) %= mod;
    }

    ans2 = 0;
    for (int k = 0; 1LL << k <= sum[n]; ++k){
        tmp = 0;
        rep(i, 0, n) p[i] = sum[i] & ((1LL << k + 1) - 1);
        sort(p, p + n + 1);
        memset(c, 0, sizeof c);
        rep(i, 0, n){
            LL now = sum[i] & ((1LL << k + 1) - 1);
            update(idx(now));
            tmp ^= query(idx(now - (1LL << k))) ^ query(idx(now + (1LL << k))) ^ query(idx(now));
        }
        if (tmp) ans2 |= 1LL << k;
    }
    

    printf("%lld %lld
", ans1 % mod, ans2);    
    return 0;

}