Educational Codeforces Round 32
2017-11-10 20:05
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C. K-Dominant Character
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given a string s consisting of lowercase Latin letters. Character c is
called k-dominant iff each substring of s with
length at least kcontains this character c.
You have to find minimum k such that there exists at least one k-dominant
character.
Input
The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 100000)
4000
.
Output
Print one number — the minimum value of k such that there exists at least one k-dominant
character.
Examples
input
output
input
output
input
output
很好的 二分 宛如智障的我 唉
#include<bits/stdc++.h>
using namespace std;
#define maxn 100000+10
char a[maxn];
int len;
int check(int x){
int c[26+10];
memset(c,0,sizeof(c));
for(int j=0;j<x;j++){
c[a[j]-'a']++;
}
int sum=0;
for(int j=0;j<26;j++){
if(c[j]) sum++;
else c[j]=-1;
}
for(int j=x;j<len;j++){
if(c[a[j]-'a']>0){
c[a[j]-'a']++;
}
c[a[j-x]-'a']--;
if(c[a[j-x]-'a']==0){
sum--;
}
if(sum==0) return 0;
}
return 1;
}
int main(){
cin>>a;
len=strlen(a);
int l=0,r=len;
while(l<r){
int mid=(l+r)/2;
if(check(mid)){
r=mid;
}
else {
l=mid+1;
}
}
cout<<l<<endl;
return 0;
}
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given a string s consisting of lowercase Latin letters. Character c is
called k-dominant iff each substring of s with
length at least kcontains this character c.
You have to find minimum k such that there exists at least one k-dominant
character.
Input
The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 100000)
4000
.
Output
Print one number — the minimum value of k such that there exists at least one k-dominant
character.
Examples
input
abacaba
output
2
input
zzzzz
output
1
input
abcde
output
3
很好的 二分 宛如智障的我 唉
#include<bits/stdc++.h>
using namespace std;
#define maxn 100000+10
char a[maxn];
int len;
int check(int x){
int c[26+10];
memset(c,0,sizeof(c));
for(int j=0;j<x;j++){
c[a[j]-'a']++;
}
int sum=0;
for(int j=0;j<26;j++){
if(c[j]) sum++;
else c[j]=-1;
}
for(int j=x;j<len;j++){
if(c[a[j]-'a']>0){
c[a[j]-'a']++;
}
c[a[j-x]-'a']--;
if(c[a[j-x]-'a']==0){
sum--;
}
if(sum==0) return 0;
}
return 1;
}
int main(){
cin>>a;
len=strlen(a);
int l=0,r=len;
while(l<r){
int mid=(l+r)/2;
if(check(mid)){
r=mid;
}
else {
l=mid+1;
}
}
cout<<l<<endl;
return 0;
}
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