題目:相似的字串
題解:這個答案滿足單調性,所以可以用二分寫。用Hash來判斷是否有相同的子串,對於其中可能有重疊的部分,用map將字串相同的進行比較。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int>P;
const int N = 2e5+10;
const ll inf = 1000000000000000000;
int base = 131;
ll Hash[N],po[N];
char s[N];
int n,k;
bool judge(int x){
map<ll,P>vis;
for(int i = x;i <= n;i++){
ll t = Hash[i]-Hash[i-x]*po[x];//可以用來判斷一個字符串是否有相同的字串
if(i-x >= vis[t].first){
vis[t].first = i;
vis[t].second++;
}
if(vis[t].second >= k) return true;
}
return false;
}
int main(){
//freopen("in.txt","r",stdin);
scanf("%d%d",&n,&k);
scanf("%s",s+1);
po[0] = 1ll;
for(int i = 1;i <= n;i++){
Hash[i] = Hash[i-1]*base+s[i]-'a';
po[i] = po[i-1]*base;
}
int l = 1,r = n;
int ans = 0;
while(l <= r){
int mid = (l+r)/2;
if(judge(mid)){
ans = mid;
l = mid+1;
}else {
r = mid-1;
}
}
cout<<ans<<endl;
return 0;
}