F. Equal XOR Segments

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2024-04-17

26.F. Equal XOR Segments2024-05-04

27.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 Game2024-06-0834.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.如果能分成偶数个区间，那么一定能分为两个区间
2.如果能分为奇数个区间，那么一定能分为三个区间
3.能分为两个区间，说明区间异或和为 $0$
4.能分为三个区间，这三个区间分别为区间 $a,b,c$ ，则 $ab$ 区间异或和为零， $bc$ 区间异或和为零

code

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

const int MAXN = 200005;
int pre[MAXN] = {0};


int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
        int n, q;
        cin >> n >> q;
        map<int,vector<int> > id;
        id[0].push_back(0);
        for (int i = 1; i <= n; i++)
        {
            int x;
            cin>>x;
            pre[i]=pre[i-1]^x;
            id[pre[i]].push_back(i);
        }
        while(q--)
        {
            int l,r;
            cin>>l>>r;
            if(pre[r]==pre[l-1]) cout<<"YES\n";
            else
            {
                int itl=*(lower_bound(id[pre[l-1]].begin(),id[pre[l-1]].end(),r)-1);//如果[l,r]区间异或和为零，代表前缀和pre[l-1]==pre[r]
                int itr=*(lower_bound(id[pre[r]].begin(),id[pre[r]].end(),l));
                //printf("itl:%d  itr:%d\n",itl,itr);
                if(itl>=itr&&itl!=l&&itr!=r) cout<<"YES\n";
                else cout<<"NO\n";
            }
        }
        cout<<"\n";
    }
    return 0;
}