CF1368B

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论审题的重要性

题面:https://codeforces.com/problemset/problem/1368/B

大致题意:规定字符串中为codeforces的不连续子串个数为k，给定数字，求最短的、子串个数至少为k的字符串

题解:易知，字符串长度为n时，尽可能平均分配即可使子串个数最大，故找到最小的n满足题设条件即可

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#include<algorithm>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<vector>
#include<string>
#include<map>
#include<cstdio>
#include<list>
#include<stack>
#include<unordered_set>
using namespace std;
using ll = int64_t;
ll T, n, m, k;
ll tmp, a[300005];
int main()
{
    T = 1;
    //cin >> T;
    while(T--)
    {
        cin >> k;
        if(k==1)
        {
            cout << "codeforces\n";
        }else
        {
            ll pos = 1;
        for (ll i = 1;;i++)
        {
            ll tem = 1;
            for (ll j = 1; j <= 10;j++)
            {
                tem *= i;
            }
            if(tem>=k)
            {
                pos = i;
                break;
            }
        }
        ll sum = 1;
        for (ll i = 1; i <= 10;i++)
        {
            a[i] = pos - 1;
            sum *= a[i];
        }
        string s = "codeforces";
        for (ll i = 1; i <= 10;i++)
        {
            if(sum<k)
            {
                a[i]++;
                sum = sum / (a[i] - 1) * a[i];
            }
            for (ll j = 1; j <= a[i];j++)
            {
                cout << s[i - 1];
            }
        }
        cout << "\n";
        }
        
    }
    return 0;
}

注意k=1时要特判