连续整数和

#include <stdio.h>                                                                                                  

void show(unsigned int value, unsigned int start, unsigned seq_end)
{
    int i = 0;
    printf("%d = %d + ", value, start);
    for(i = 1; i < seq_end; i++)
    {
        printf("%d + ", start + i);
    }
    printf("%d\n", start + seq_end);
}

/*
 * 功能：连续正整数和测试
 *       若存在则输出相应的等式
 *
 * 返回：返回存在的组数，不存在则返回0
 */
int integer_solve(unsigned value)
{
    unsigned int start = 0;
    unsigned int all_group = 0;
    if(!(value &(value-1)))//2^n必定不满足
    {
        return 0;
    }
    for(start = 1; start <= (value/2);start++)//起始值必小于(value/2);
    {
        unsigned int count_start = 0;
        unsigned int count_end = (value+1)/2 - start;//终值必小于(value+1)/2，获得最多测试项数
        unsigned int sum = 0;
        for(count_start = 1; count_start <= count_end; count_start++)
        {
            sum = (2 * start + count_start) * (count_start + 1)/2;
            if(sum >= value)
                break;
            sum = 0;
        }
        if(sum)
        {
            if(sum == value)
            {
                all_group++;
                show(value, start, count_start);
            }
        }
    }
    return all_group;
}

int main(void)
{
    int i = 0;
    for(i = 0; i <= 100; i++)
    {
        unsigned int all_group = 0;
        all_group = integer_solve(i);
        if(all_group)
        {
            printf("********************************\n");
            printf("********************************\n");
        }
    }
    return 0;