C语言程序设计基础01 计算分数不等式

【题目来源】                                                                                    

　　《趣味C程序设计集锦》p4

【题目分析】                                                                                  

　　简单题......

　　s += 1.0/++i;

【题目代码】                                                                                  

/*============================================================================*\
* 计算 < 1 + 1/2 + 1/3 + ··· + 1/m < n+1
* @date 3/12/2013
* VC++ 6.0
\*============================================================================*/
#include <stdio.h>
#include <stdlib.h>

int main()
{
    long c, d , i = 0, n;
    double s = 0.0;
    printf("计算 < 1 + 1/2 + 1/3 + ··· + 1/m < n+1\n");
    printf("\n请输入n:");
    scanf("%ld", &n);

    while(s < n){
        s += 1.0/++i;
    }

    c = i;

    while(s < n+1){
        s += 1.0/++i;
    }

    d = i-1;

    printf("\n满足不等式的m为：%ld ≤ m ≤ %ld \n", c,d);

    return 0;
}

【测试结果】