C. Did We Get Everything Covered

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 Problemset2024-01-28

38.C. Did We Get Everything Covered2024-01-28

39.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

前情提要

限于自身知识水平的储备不足，无法对这道题的贪心算法做出一个证明，待来日学识渐长把这个证明写下

题解

我们可以把字符串s分成若干区间，每一区间对应一位数字的储备
已知长度为n，那我们就一位一位地遍历，一旦所有元素遍历齐就开始下一位的遍历，因为再往后遍历也不起作用
然后取每一位最后一个收集齐的字母因为这一位想要出现这个字母就必须要取它，最后几位取没有遍历到的字母重复即可

code

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n,k,m;
        cin>>n>>k>>m;
        string s;
        cin>>s;

        int a[27]={0},cnt=0;
        string ans="";
        for(int i=0;s[i];i++)
        {
            if(!a[s[i]-97])cnt++;
            a[s[i]-97]=1;
            if(cnt==k)
            {
                ans+=s[i];
                cnt=0;
                memset(a,0,sizeof(a));
            }
        }
        if(ans.size()>=n)puts("YES");
        else
        {
            puts("NO");
            for(int i=0;i<k;i++)
            {
                if(!a[i])
                {
                    while(ans.size()<n)ans+=('a'+i);
                }
            }
            cout<<ans<<endl;
        }
    }
    return 0;
}