D. A BIT of an Inequality

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 Segments2024-05-04

27.D. A BIT of an Inequality2024-05-04

28.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 Game2024-06-0834.D. Sum of XOR Functions2024-06-1435.C. Medium Design2024-06-2836.F. Feed Cats2024-07-0737.E. Boring Segments2024-07-28

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原题链接

题解

1.如果 $x\oplus y>x$ ，则 $y$ 的最高位对应的 $x$ 一定是 $0$
2.$f(x,y)\oplus f(y,z)>f(x,z)$ 等价于 $f(x,z)\oplus a_y>f(x,z)$
3.$x\in [1,y]$ 则 $x-1\in [0,y-1]$

code

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

#define ll long long

ll a[100005]={0}, pre[100005]={0};
ll cntleft[100005][35][2]={0}, cntright[100005][35][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 t;
    read(t);
    while(t--)
    {
        ll n;
        read(n);
        for(ll i=1; i<=n; i++)
        {
            read(a[i]);
            pre[i]=pre[i-1]^a[i];
        }

        for(ll k=0; k<=30; k++) cntleft[0][k][0]=1;
        for(ll i=1; i<=n; i++)
        {
            for(ll k=0; k<=30; k++)
            {
                ll add=((pre[i]>>k)&1);
                cntleft[i][k][add]=cntleft[i-1][k][add]+1;
                cntleft[i][k][1-add]=cntleft[i-1][k][1-add];
            }
        }

        for(ll k=0; k<=30; k++) cntright[n+1][k][1]=cntright[n+1][k][0]=0;
        for(ll i=n; i>=1; i--)
        {
            for(ll k=0; k<=30; k++)
            {
                ll add=((pre[i]>>k)&1);
                cntright[i][k][add]=cntright[i+1][k][add]+1;
                cntright[i][k][1-add]=cntright[i+1][k][1-add];
            }
        }

        ll ans=0;
        for(ll i=1; i<=n; i++)
        {
            ll height=log2(a[i]);
            ans+=cntleft[i-1][height][0]*cntright[i][height][0];
            ans+=cntleft[i-1][height][1]*cntright[i][height][1];
        }
        write(ans);
        putchar('\n');
    }
    return 0;
}