1157: n consecutive ones

1157: Successive n individual 1

Title Description
Computer data is made up of 0 and 1 Composed of , Look at the long 0101001110101111011, To find continuity n individual 1 How many substrings are there , It's a real hassle , Please program it .
Input
Enter a string for the first line , from 0 and 1 form , The length is less than 1000; Enter a positive integer for the second line n.
Output
Output an integer , Means continuous n One of the 1 The number of substrings of .
The sample input Copy
0101001110101111011
2
Sample output Copy
6
source / classification

#include<stdio.h>
#include<string.h>
#define MAX 1000


int main(){
    
    char s[MAX];
    int n,f=0,count=0;
    scanf("%s%d",&s,&n);

    // Pay attention to the interval [0-strlen(s)-n]
    for(int i=0;i<=strlen(s)-n;i++){
    
        if(s[i]=='1'){
    
            f=1;
            // Satisfy (i-i+n) common n individual 1,f=1
            for(int j=i;j<i+n;j++){
    
                if(s[j]=='0') {
    
                    f=0;
                    break;
                }
            }
            if(f==1) count++;
        }
    }
    printf("%d\n",count);
    return 0;
}