1147: find subarray

1147: Find subarray

Title Description
Given two arrays of integers , Array a Yes n Elements , Array b Yes m Elements ,1<=m<=n<100, Please check the array b Is it an array a Subarray . If from array a Some element of a[i] Start , Yes b[0]=a[i],b[1]=a[i+1],…,b[m]=a[i+m], It's called an array b It's an array a Subarray .
Input
Enter two integers in the first line n and m; Second row array a Of n It's an integer ; The third line is array b Of m It's an integer , The data are separated by spaces .
Output
Output takes up one line . if b yes a Subarray , Then the position of the output subarray i, Pay attention to subscripts from 0 Start ; Otherwise output “No Answer”.
The sample input Copy
8 3
3 2 6 7 8 3 2 5
3 2 5
Sample output Copy
5
source / classification

#include<stdio.h>
#include<string.h>
#define N 100
/*  Find subarray   First find the same subscript in the array as the first bit of the subarray i, Then compare   If it's different later , Then from i Start looking for  */
int main(){
    
    int n,m,k,f,c=0;
    int a[N],b[N];
    scanf("%d %d",&n,&m);
    for(int i=0;i<n;i++){
    
        scanf("%d",&a[i]);
    }
    for(int i=0;i<m;i++){
    
        scanf("%d",&b[i]);
    }


    for(int i=0;i<n;i++){
    
        f=1;
        // Find the first identical subscript 
        if(b[0]==a[i]) {
    
        	// Start sequential comparison a,b Array , If not, exit the cycle immediately 
            for(int j=1;j<m;j++){
    
                if(b[j]!=a[i+j]){
    
                    f=0;
                    break;
                }
            }
            // There may be a variety of situations ,c Used to record whether there are subarrays 
            if(f==1) {
    
                c=1;
                printf("%d\n",i);
            }
        }

    }
    if(c==0) printf("No Answer\n");
	return 0;
}