c - 给定一个正整数，判断它是否等于四个连续整数之和

Question

问题:

Given a positive integer, tell whether it is equal to the sum of four consecutive integers.

Input Format:
Input consists of a single positive integer,

Output Format:
Output consists of a string which is either "yes" or "no". Print yes when the given number is the sum of four consecutive integers and print no otherwise.

Example:
In this example, 2 = -1 + 0 + 1 + 2, and 82 = 19 + 20 + 21 + 22. 5 and 41 are not of the correct form.

Sample Input 1:
2

Sample Output 1:
yes

Sample Input 2:
41

Sample Output 2:
no

我的程序未被接受，并且对于某些值，我的结果正在重复，任何人都可以帮助我吗？

#include <stdio.h>

int main()
{
    int num,sum1=0,index,sum2=0,Soln=0;
    int temp,i;
    scanf("%d",&num);
    if(num <= 0 )
    {
        printf("Wrong Input\n");
        return 0;
    }
    for(index=1;;index++)
    {
        sum1=sum1+index;
        if(sum1>=num)
            break;
    }
    if(sum1 == num )
    {
        if(num == index)
        {
            printf("yes");
            return 0;
        }
        Soln++;
        printf("no" );
        for(temp=1;temp<=index;temp++)
            printf("no");
    }
    for(i=1;i<index;i++)
    {
        sum2+=i;
        if((sum1-sum2) == num )
        {
            if((i+1) == index)
            {
//printf("yes");
                break;
            }
            Soln++;
            printf("no" );
            for(temp= i+1;temp<=index;temp++)
                printf("no");
            break;
        }
    }
    if( Soln == 0 && num >= 9 && num/2 == (num-1)/2 ) //--> if U don't want more       than one ,soln use this 
    //if( num >= 9 && num/2 == (num-1)/2 ) //--> if U want more than one soln, use this
    {
        Soln++;
        printf("no");
    }
    if(Soln==0)
    {
        printf("yes");
    }
    return 0 ;
}

Answer 1

取四个连续的整数 k、k+1、k+2 和 k+3 并将它们相加。得出 4k + 6。如果整数 n 是四个连续整数的和，则意味着对于某个整数 k，n = 4k + 6。等价地，这意味着对于某个整数 k，n - 6 = 4k，或者等价地 (n - 6) mod 4 = 0。这可能会帮助您显着简化您的程序。