CodeForces - 849C From Y to Y


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For a given unordered multiset of n lowercase English letters (“multi” means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:

Remove any two elements s and t from the set, and add their concatenation s + t to the set. The cost of such operation is defined to be , where f(s, c) denotes the number of times character c appears in string s.

Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.

Input The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.

Output Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.

Note that the printed string doesn’t need to be the final concatenated string. It only needs to represent an unordered multiset of letters.

Examples Input 12 Output abababab Input 3 Output codeforces Note For the multiset {‘a’, ‘b’, ‘a’, ‘b’, ‘a’, ‘b’, ‘a’, ‘b’}, one of the ways to complete the process is as follows:

{“ab”, “a”, “b”, “a”, “b”, “a”, “b”}, with a cost of 0; {“aba”, “b”, “a”, “b”, “a”, “b”}, with a cost of 1; {“abab”, “a”, “b”, “a”, “b”}, with a cost of 1; {“abab”, “ab”, “a”, “b”}, with a cost of 0; {“abab”, “aba”, “b”}, with a cost of 1; {“abab”, “abab”}, with a cost of 1; {“abababab”}, with a cost of 8. The total cost is 12, and it can be proved to be the minimum cost of the process.


题意难读懂的地方在于那个计算 在这里插入图片描述

这个公式的意思是 两个子串s和t c在a到z里面取,每个都取一遍, s里字母c的数量和t里字母c的数量的积, 把c在a到z里的每一个结果相加

举例: ‘aab’和’abb’代表着s和t 从a到z依次选择c

c = a时: aab里面有两个a,abb里面有一个a,那么结果就是21 = 2 c = b时: aab里面有一个b,abb里面有两个b,那么结果就是12 = 2 c = 字符c时: aab和abb里面没有c,结果为0*0 = 0 其他依次下去都是0 结果为4

然后,观察结果 a 0 aa 1 aaa 3 aaaa 6 aaaaa 10 … 可以看出规律, i*(i-1)/2




using namespace std;
const int MAXN = 1e4;
int arr[MAXN];
int main(){
    for(int i=1;i<MAXN;i++) arr[i]=arr[i-1]+i; // 先打表出来
    int k;cin>>k;
    char temp='a';
    if(k==0) putchar(temp);//特判一下
        int cnt=lower_bound(arr,arr+MAXN,k)-arr;
        if(arr[cnt]>k) cnt--;// 找到第一个小于k的值
        cnt++;  // 因为序号为0的时候,是1个a
        while(cnt--) putchar(temp);
        temp++;// 改变字符
    return 0;