D. Sum of XOR Functions

合集 - 区间问题(37)

1.Iva & Pav2024-02-172.P2023 [AHOI2009] 维护序列2024-02-203.P1531 I Hate It2024-02-204.P5057 [CQOI2006] 简单题2024-02-205.P2357 守墓人2024-02-206.P8775 [蓝桥杯 2022 省 A] 青蛙过河2024-02-217.Dora and Search2024-02-248.P6492 [COCI2010-2011#6] STEP2024-02-269.D. Slimes2024-02-2810.P1040 [NOIP2003 提高组] 加分二叉树2024-03-0311.P2466 [SDOI2008] Sue 的小球2024-03-0412.P3957 [NOIP2017 普及组] 跳房子2024-03-0613.P2135 方块消除2024-03-0614.P1884 [USACO12FEB] Overplanting S2024-03-1015.P2642 双子序列最大和2024-03-2516.P2422 良好的感觉2024-03-2517.P3128 [USACO15DEC] Max Flow P2024-04-0218.P3258 [JLOI2014] 松鼠的新家2024-04-0219.P3384 【模板】重链剖分/树链剖分2024-04-0320.P5367 【模板】康托展开2024-04-0521.P2344 [USACO11FEB] Generic Cow Protests G2024-04-1322.P2709 小B的询问2024-04-1623.P4145 上帝造题的七分钟 2 / 花神游历各国2024-04-1724.P5677 [GZOI2017] 配对统计2024-04-1725.P2161 [SHOI2009] 会场预约2024-04-1726.F. Equal XOR Segments2024-05-0427.D. A BIT of an Inequality2024-05-0428.P3147 [USACO16OPEN] 262144 P2024-05-1329.P4290 [HAOI2008] 玩具取名2024-05-1330.E. Long Inversions2024-04-0931.P1668 [USACO04DEC] Cleaning Shifts S2024-05-1732.P9691 [GDCPC2023] Base Station Construction2024-05-1833.P2734 [USACO3.3] 游戏 A Game2024-06-08

34.D. Sum of XOR Functions2024-06-14

35.C. Medium Design2024-06-2836.F. Feed Cats2024-07-0737.E. Boring Segments2024-07-28

收起

原题链接

题解

1.暴力枚举每一个区间，然后加和 $\to O(n^2)$
如何优化？考虑到区间异或和不一定每一位都对答案有贡献，所以我们只考虑对答案有贡献的区间
2.遍历每一位，找出能使他对答案有贡献的区间个数，再乘上区间长度

细节

由于有求模运算，所以减法可能会出现负数，通过加一个模数解决

code

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll mod=998244353;
ll a[300005];
ll sum[2]={0};
ll cnt[2]={0};

inline void read(ll &x) {
    x = 0;
    ll flag = 1;
    char c = getchar();
    while(c < '0' || c > '9'){
        if(c == '-') flag = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = (x << 3) + (x << 1) + (c ^ 48);
        c = getchar();
    }
    x *= flag;
}

inline void write(ll x)
{
    if(x < 0){
        putchar('-');
        x = -x;
    }
    if(x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
}

int main()
{
    ll n;
    read(n);

    for(ll i=1; i<=n; i++) read(a[i]);

    ll ans=0;
    for(ll k=0; k<=31; k++)
    {
        ll flag=0;
        sum[0]=0;
        sum[1]=0;
        cnt[0]=1;
        cnt[1]=0;
        for(ll i=1; i<=n; i++)
        {
            flag^=((a[i]>>k)&1LL);
            ans = (ans + (i * cnt[flag^1] % mod - sum[flag^1] + mod) % mod * (1LL << k) % mod) % mod;
            sum[flag] = (sum[flag] + i) % mod;
            cnt[flag] = (cnt[flag] + 1) % mod;
        }
    }
    write(ans);
    return 0;
}