题目
You are given a string s. Among the different substrings of s, print the K-th lexicographically smallest one.
A substring of s is a string obtained by taking out a non-empty contiguous part in s. For example, if s = ababc, a, bab and ababc are substrings of s, while ac, z and an empty string are not. Also, we say that substrings are different when they are different as strings.
Let X=x1x2…xn and Y=y1y2…ym be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or xj>yj where j is the smallest integer such that xj≠yj.
Constraints
- 1 ≤ |s| ≤ 5000
- s consists of lowercase English letters.
- 1 ≤ K ≤ 5
- s has at least K different substrings.
Partial Score
- 200 points will be awarded as a partial score for passing the test set satisfying |s| ≤ 50.
Input
Input is given from Standard Input in the following format:
s
K
Output
Print the K-th lexicographically smallest substring of K.
题目大意
给你一个字符串s，求它字典序第k大的字串
分析
因为k<=5，所以我们可以枚举字串的起始点和长度，以为长度小，我们把整个字串转化成数字然后排序即可，注意长度不满k的字串在末尾要补零
代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<ctime>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
long long a[110000];
map<long long,bool>app;
int cnt;
long long go(string s,int le,int ri){
int i,j,k;
long long res=0;
for(i=le;i<=ri;i++)
res=res*100+(s[i]-'a'+1);
for(i=ri-le+2;i<=5;i++)
res*=100;
return res;
}
int main()
{ int n,m,i,j,k;
string s;
cin>>s;
n=s.length();
cin>>k;
for(i=0;i<n;i++)
for(j=i;j<=min(i+k,n-1);j++){
long long res=go(s,i,j);
if(!app[res]){
a[++cnt]=res;
app[res]=1;
}
}
sort(a+1,a+cnt+1);
char c[110];
cnt=0;
while(a[k]){
if(a[k]%100)c[++cnt]=(a[k]%100-1)+'a';
a[k]/=100;
}
for(i=cnt;i>0;i--)
cout<<c[i];
return 0;
}