C. Removal of Unattractive Pairs

合集 - codeforces(57)

1.C. Insert and Equalize2023-12-06

2.C. Removal of Unattractive Pairs2023-12-06

3.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!2024-02-2851.C. Find B2024-02-2852.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

收起

原题链接

不知道这个思想叫什么，应该叫结果思想

导论
如果存在一个最长的字符串，我又没有可能把他消掉？
答案是，只要其他字符的长度大于等于最长字符串的长度，就一定能把他消掉。
所以我们不考虑字符串是怎么消除的，直接看结果。

原因解释如下
1.该最长字符串一定和其他字符相连，则消除操作显然。
2.如果在执行某个消除操作后遇到了与最长字符串同一字符，则更新最长字符串，情况又回到了1。
3.执行多个2后，这道题就成了找一个字符出现的最多次数。

特判考虑
1.如果n为奇数，至少剩一个字符
2.设最长字符串（即一个字符出现的最多次数）为len，如果len<=n/2，则能全部消掉，但如果是奇数还会剩一个
3.如果2*len>n 则一定剩2*len-n个字符

代码

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        char s[1000005];
        scanf("%d%s",&n,s);
        int a[30]={0};
        for(int i=0;s[i];i++)a[s[i]-97]++;
        int len=0;
        for(int i=0;i<26;i++)len=max(len,a[i]);
        if(2*len<=n)   printf("%d\n",n%2==0?0:1);
        else printf("%d\n",2*len-n);
    }
    return 0;
}