Codeforces Round #431 (Div. 2) C
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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 cappears 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.
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
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.
12
abababab
3
codeforces
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:可能说的不清楚,我们取两个字符串,重复的我们把出现次数记录一下,然后相乘
a和b没有重复的,0*0
aba和b有一个重复的 1*1
然后。。为什么最后等于8了我也没想(为什么不是4*4或者1*1?)
反正最后我们加起来等于n就行
解法:
1 构造当然想最容易的 n=12
a a a a这种合并就很简单,0+1+2+3就行
2 我们拿5个a,花费了10,还差2
3 换个字母b,拿两个b b ,还差1
4 再换个字母c,两个c c 搞定
#include<bits/stdc++.h>
using namespace std;
double x[];
set<double>Se;
double ans;
int main(){
int n;
cin>>n;
string s="";
if(n==){
cout<<"a"<<endl;
}else{
char c='a';
while(n){
int sum=;
int i=;
for(i=;sum<=n;i++){
sum+=i;
} n-=(sum-i+);
for(int j=;j<i-;j++){
s+=c;
}
c++;
}
cout<<s<<endl;
}
return ;
}
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