题目
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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.
解释
题意难读懂的地方在于那个计算
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
如果从另一个字母开始,就重新开始计数
我们需要找到第一个小于k的值的数,然后从那个数开始用另一个字母重新开始计数
代码
#include<bits/stdc++.h>
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
while(k){
int cnt=lower_bound(arr,arr+MAXN,k)-arr;
if(arr[cnt]>k) cnt--;// 找到第一个小于k的值
k-=arr[cnt];
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;
}