string中 find 查询

B. Segment Occurrences

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two strings $\mathrm{ss and tt,\; both\; consisting\; only\; of\; lowercase\; Latin\; letters.}$

The substring $\mathrm{s[l..r]s[l..r] is\; the\; string\; which\; is\; obtained\; by\; taking\; characters sl,sl+1,\dots ,srsl,sl+1,\dots ,sr without\; changing\; the\; order.}$

Each of the occurrences of string $\mathrm{aa in\; a\; string bb is\; a\; position ii (1\le i\le |b|-|a|+11\le i\le |b|-|a|+1)\; such\; that b[i..i+|a|-1]=ab[i..i+|a|-1]=a (|a||a| is\; the\; length\; of\; string aa).}$

You are asked $\mathrm{qq queries:\; for\; the ii-th\; query\; you\; are\; required\; to\; calculate\; the\; number\; of\; occurrences\; of\; string tt in\; a\; substring s[li..ri]s[li..ri].}$

Input

The first line contains three integer numbers $\mathrm{nn, mm and qq (1\le n,m\le 1031\le n,m\le 103, 1\le q\le 1051\le q\le 105)\; —\; the\; length\; of\; string ss,\; the\; length\; of\; string ttand\; the\; number\; of\; queries,\; respectively.}$

The second line is a string $\mathrm{ss (|s|=n|s|=n),\; consisting\; only\; of\; lowercase\; Latin\; letters.}$

The third line is a string $\mathrm{tt (|t|=m|t|=m),\; consisting\; only\; of\; lowercase\; Latin\; letters.}$

Each of the next $\mathrm{qq lines\; contains\; two\; integer\; numbers lili and riri (1\le li\le ri\le n1\le li\le ri\le n)\; —\; the\; arguments\; for\; the ii-th\; query.}$

Output

Print $\mathrm{qq lines\; —\; the ii-th\; line\; should\; contain\; the\; answer\; to\; the ii-th\; query,\; that\; is\; the\; number\; of\; occurrences\; of\; string tt in\; a\; substring s[li..ri]s[li..ri].}$

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
typedef long long ll;

int main()
{
    int a,b,c,l,r,i;
    string d,e;
    scanf("%d%d%d",&a,&b,&c);
    cin>>d,cin>>e;
    while(1)
    {
        int p=d.find(e);
        if(p!=-1)
            d[p]='1';
        else
            break;
    }
    while(c--)
    {
        int ans=0;
        cin>>l>>r;
        l--,r--;
        for(i=l;i<=r+1-b;i++)
        {
            if(d[i]=='1')
                ans++;
        }
        printf("%d\n",ans);



    }

    return 0;
}

 

 

转载于:https://www.cnblogs.com/bhd123/p/9636693.html