Educational Codeforces Round 67 (Rated for Div. 2)B. Letters Shop
题意:找到从头开始最短的串,使得串的字母个数涵盖给出子串所有字母的个数
做法:一开始直接暴力计数加查找,后来TLE了才想到用二分。。。我真的是傻了
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<queue>
#include<map>
#define ll long long
#define pb push_back
#define rep(x,a,b) for (int x=a;x<=b;x++)
#define repp(x,a,b) for (int x=a;x<b;x++)
#define W(x) printf("%d\n",x)
#define WW(x) printf("%lld\n",x)
#define pi 3.14159265358979323846
#define mem(a,x) memset(a,x,sizeof a)
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
using namespace std;
const int maxn=2e5+7;
const int INF=1e9;
const ll INFF=1e18;
char s[maxn];
char s1[maxn];
int num[27][maxn];//记录原始串在第i个位置之前各个字母的数量
int num1[27];//记录所求串的字母个数
int main()
{
int l,m;
scanf("%d",&l);
scanf("%s",s);
rep(i,0,l-1)//打表
{
repp(j,0,26)
if (i!=0)num[j][i]=num[j][i-1];
num[s[i]-'a'][i]++;
}
scanf("%d",&m);
while(m--)
{
scanf("%s",s1);
rep(i,0,26)//记得清空
num1[i]=0;
int l1=strlen(s1);
repp(i,0,l1)
num1[s1[i]-'a']++;
int ans=0;
repp(i,0,26)
{
int left=0,r=l-1,mid;
if (num1[i]==0)continue;
while(left<r)
{
mid=(left+r)>>1;
if (num[i][mid]>=num1[i])r=mid;
else left=mid+1;
}
ans=max(ans,left+1);
}
W(ans);
}
return 0;
}