打印给定字符串中以 1 开头和结尾的子字符串的数量

Question

For example, if the input string is “00100101”, then the output substrings should be “1001”, “100101” and “101”.

i think there is a mistake with function but not sure. I got 2 errors Line 24, col 6 [Error] conflicting types for 'binarysubstring' and Line 4, col 6 [Note] previous declaration of 'binarysubstring' was here

#include<stdio.h>
#include<string.h>

long binarysubstring(int, char);

int main()
{
    int t,n,count;
    long r;
    char a[n+1];
    t=0;
    scanf("%d",&n);
    scanf("%s",&a);
    r=binarysubstring(n, a[n]);
    printf("%ld",r);
}

long binarysubstring(int n,char a[n])
{
    int i;
    long r;
    for(i=0;i<n;i++)
    {
        if(a[i]==1)
        {
            r+=1;
        }
    }
    r=((r*(r-1))/2);
    return r;
}

Answer 1

As there are multiple mistakes to process here, I'll try to explain and fix them starting from the errors you see on compilation.

First, your compilation error is because your function declaration takes a char as the second argument instead of a character array , line 4 should instead be

long binarysubstring(int, char[]);

You also should not be declaring the array size inside the function signature on line 18.

Your other immediate issue is that you initialize the buffer char a[n+1]; on line 10 before the input to define n on line 12, and since you did not define a value when n was declared on line 8, line 10 will actually declare a character array of size n + 1 = 0 + 1 = 1, which cannot store any input.

With these three things fixed, the following code compiled for me with Clang-7:

#include<stdio.h>
#include<string.h>

long binarysubstring(int, char[]); // Added the brackets here

int main()
{
    int t,n,count;
    long r;
    t=0;
    scanf("%d",&n); // Moved this...
    char a[n+1]; // ...to be before this
    scanf("%s", a);
    r=binarysubstring(n, a);
    printf("%ld",r);
}

long binarysubstring(int n, char a[/*Removed n, do not pass anything here*/]) 
{
    int i;
    long r;
    for(i=0;i<n;i++)
    {
        if(a[i]==1)
        {
            r+=1;
        }
    }
    r=((r*(r-1))/2);
    return r;
}

However, this code always returns 0 because on line 24, your if statement compares each char to an integer , which is parsed to its ASCII code that is not the actual character of 1. As a result, r is never incremented because the condition is never true with the binary strings you are inputting, and it remains 0 up to your final sum formula, which then evaluates to 0 as well.

At this point, I'll give my attempt to implement what your question asks:

#include <stdio.h>

long binarysubstring(unsigned short, char[]);

int main() {
    unsigned short n;
    printf("Input the length of your string: ");
    scanf("%hu", &n);
    char input[n + 1];
    printf("Input the binary string: ");
    scanf("%s", input);
    long answer = binarysubstring(n, input);
    printf("Answer: %ld\n", answer);
}

long binarysubstring(unsigned short n, char a[]) {
    long r = 0;
    for (unsigned short i = 0; i < n; ++i) {
      if (a[i] == '1') {
        ++r;
      }
    }
    return r*(r-1)/2;
}

This assumes the input is a binary string. I use the same algorithm you appear to, counting the number of times 1 appears in the string with r and then returning the sum of the integers in the sequence from 1 to r , inclusive. It also happens to return 0 for binary strings of length 1.