codeforces 1016B.Segment Occurrences

B. Segment Occurrence

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

$\mathrm{题目大意:输入两个长度分别为n,m的字符串a,b，求在a中指定区域[l,r]上有多少个b。}$

$\mathrm{思路：在a中查找b，每个开头位置设置标记，在给定区域查找标记即可。}$

$\mathrm{代码：}$

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <algorithm>
 4 #include <iostream>
 5 using namespace std;
 6 typedef long long ll;
 7 
 8 int main()
 9 {
10     int n,m,k;
11     string a,b;
12     while(scanf("%d%d%d",&n,&m,&k)==3)
13     {
14         cin>>a;
15         cin>>b;
16         while(1)
17         {
18             int p=a.find(b);
19             if(p!=-1)
20             {
21                 a[p]='1';
22             }
23             else
24                 break;
25         }
26         while(k--)
27         {
28             int l,r,ans=0;
29             cin>>l>>r;
30             l--;
31             r--;
32             for(int i=l; i<=r-m+1; i++)
33             {
34                 if(a[i]=='1')
35                     ans++;
36             }
37             printf("%d\n",ans);
38         }
39     }
40     return 0;
41 }

 

转载于:https://www.cnblogs.com/wangxwws/p/9503833.html