C. Find B

合集 - codeforces(57)

1.C. Insert and Equalize2023-12-062.C. Removal of Unattractive Pairs2023-12-063.D. Jumping Through Segments2023-12-084.E. Good Triples2023-12-085.F. Shift and Reverse2023-12-086.D. Yet Another Monster Fight2023-12-117.A. Constructive Problems2023-12-188.C. Game with Multiset2023-12-199.A. Rating Increase2023-12-1910.B. Swap and Delete2023-12-1911.A. Problemsolving Log2023-12-2012.B. Preparing for the Contest2023-12-2013.C. Quests2023-12-2014.D. Three Activities2023-12-2015.E2. Game with Marbles (Hard Version)2023-12-2016.cf刷题有感2023-12-2017.A. Anonymous Informant2023-12-2118.A. Forked!2023-12-2319.B. Make Almost Equal With Mod2023-12-2720.C. Heavy Intervals2023-12-2721.D. Split Plus K2023-12-2822.A. 20232023-12-3123.B. Two Divisors2023-12-3124.C. Training Before the Olympiad2023-12-3125.D. Mathematical Problem2024-01-0226.F. Greetings2024-01-0227.C. Partitioning the Array2024-01-1528.G. Bicycles2024-01-1629.E. Eat the Chip2024-01-1730.G. Lights2024-01-1731.D. Array Repetition2024-01-1732.D. Berserk Monsters2024-01-2233.E. Increasing Subsequences2024-01-2334.D. Very Different Array2024-01-2435.G. Mischievous Shooter2024-01-2536.B. Plus-Minus Split2024-01-2737.B. A Balanced Problemset2024-01-2838.C. Did We Get Everything Covered2024-01-2839.D. Find the Different Ones!2024-02-0840.C. Grouping Increases2024-02-1041.D. Good Trip2024-02-1242.C. Physical Education Lesson2024-02-1243.E. Final Countdown2024-02-1844.D. Divisible Pairs2024-02-1945.G. Vlad and Trouble at MIT2024-02-2046.A. Brick Wall2024-02-2847.B. Minimize Inversions2024-02-2848.C. XOR-distance2024-02-2849.A. Moving Chips2024-02-2850.B. Monsters Attack!2024-02-28

51.C. Find B2024-02-28

52.D. Slimes2024-02-2853.C. Turtle Fingers: Count the Values of k2024-02-2854.D. Turtle Tenacity: Continual Mods2024-02-2855.D. Vlad and Division2024-02-2856.C. LR-remainders2024-02-2857.D. Lonely Mountain Dungeons2024-03-12

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原题链接

题解

给定数组c，和若干查询区间，请问能否改变区间中的每一个值且区间和还不变？

对于任意一个数，不是加就是减，而对于整个数组而言，加了多少就要减多少
而对于等于1的元素而言，只能加，因此我们令元素为1的为待加元素，其他元素均为待减元素
找出所有大于1的元素把他们变成一，然后差值累加到一个变量sum里
再判断1的个数k，看看sum能否补齐k的缺口，即sum>=k
如果k=0，即没有待加元素，但是指定任意一个元素为待加元素
那就判断区间大小是否大于等于2，即加和减的元素至少各有一个

code

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

ll pres[300005]={0};
ll add[300005]={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,q;
        read(n); read(q);
        for(ll i=1;i<=n;i++)
        {
            ll x;
            read(x);
            pres[i]=pres[i-1]+x-1;//区间和
            add[i]=add[i-1]+(x==1);//区间1的个数
        }
        while(q--)
        {
            ll x,y;
            read(x); read(y);
            ll k=add[y]-add[x-1],sum=pres[y]-pres[x-1];
            if(k&&sum>=k||!k&&y-x+1>=2) puts("YES");
            else puts("NO");
        }
    }
    return 0;
}