cf 502 div2 题目链接http://codeforces.com/problemset/problem/1016/B
题意 第一行第一个数字主串的长度，第二个是子串长度，第三个数字代表几次询问
例如 1-3 在主串中子串出现的次数
kmp模板题
具体kmp算法详见https://blog.csdn.net/u010232171/article/details/41945605
#include<iostream>
#include<cstdio>
#include<string>
#include<math.h>
using namespace std;
const int maxn=1005;
char s[maxn],t[maxn];
int slen,tlen;
int next[maxn];
void getnext()
{
int j,k;
j=0;k=-1;next[0]=-1;
while(j<tlen)
if(k==-1||t[j]==t[k])
next[++j]=++k;
else
k=next[k];
}
int kmp_count(int m,int n)
{
int ans=0;
int i,j=0;
if(slen==1&&tlen==1)
{
if(s[0]==t[0])
return 1;
else
return 0;
}
getnext();
for(int i=m;i<n;i++)
{
while(j>0&&s[i]!=t[j])
j=next[j];
if(s[i]==t[j])
j++;
if(j==tlen)
{
ans++;
j=next[j];
}
}return ans;
}
int main()
{
int q;
while(scanf("%d%d%d",&slen,&tlen,&q)!=EOF)
{
scanf("%s",s);
scanf("%s",t);
while(q--)
{
int m,n;
scanf("%d%d",&m,&n);
printf("%d\n",kmp_count(m-1,n));
}
}
return 0;
}
kmp模板
#include <iostream>
#include <cstring>
using namespace std;
const int N = 1000002;
int next[N];
char S[N], T[N];
int slen, tlen;
void getNext()
{
int j, k;
j = 0; k = -1; next[0] = -1;
while(j < tlen)
if(k == -1 || T[j] == T[k])
next[++j] = ++k;
else
k = next[k];
}
/*
返回模式串T在主串S中首次出现的位置
返回的位置是从0开始的。
*/
int KMP_Index()
{
int i = 0, j = 0;
getNext();
while(i < slen && j < tlen)
{
if(j == -1 || S[i] == T[j])
{
i++; j++;
}
else
j = next[j];
}
if(j == tlen)
return i - tlen;
else
return -1;
}
/*
返回模式串在主串S中出现的次数
*/
int KMP_Count()
{
int ans = 0;
int i, j = 0;
if(slen == 1 && tlen == 1)
{
if(S[0] == T[0])
return 1;
else
return 0;
}
getNext();
for(i = 0; i < slen; i++)
{
while(j > 0 && S[i] != T[j])
j = next[j];
if(S[i] == T[j])
j++;
if(j == tlen)
{
ans++;
j = next[j];
}
}
return ans;
}
int main()
{
int TT;
int i, cc;
cin>>TT;
while(TT--)
{
cin>>S>>T;
slen = strlen(S);
tlen = strlen(T);
cout<<"模式串T在主串S中首次出现的位置是: "<<KMP_Index()<<endl;
cout<<"模式串T在主串S中出现的次数为: "<<KMP_Count()<<endl;
}
return 0;
}
题目
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 ss and tt, both consisting only of lowercase Latin letters.
The substring s[l..r]s[l..r] is the string which is obtained by taking characters sl,sl+1,…,srsl,sl+1,…,sr without changing the order.
Each of the occurrences of string aa in a string bb is a position ii (1≤i≤|b|−|a|+11≤i≤|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 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 nn, mm and qq (1≤n,m≤1031≤n,m≤103, 1≤q≤1051≤q≤105) — the length of string ss, the length of string tt and the number of queries, respectively.
The second line is a string ss (|s|=n|s|=n), consisting only of lowercase Latin letters.
The third line is a string tt (|t|=m|t|=m), consisting only of lowercase Latin letters.
Each of the next qq lines contains two integer numbers lili and riri (1≤li≤ri≤n1≤li≤ri≤n) — the arguments for the ii-th query.
Output
Print 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].
Examples
input
Copy
10 3 4 codeforces for 1 3 3 10 5 6 5 7
output
Copy
0 1 0 1
input
Copy
15 2 3 abacabadabacaba ba 1 15 3 4 2 14
output
Copy
4 0 3
input
Copy
3 5 2 aaa baaab 1 3 1 1
output
Copy
0 0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.