原题如下：考虑两个无线序列：
1.0＋1.0/2.0+1.0/3.0+1.0/4.0+……
1.0-1.0/2.0+1.0/3.0-1.0/4.0+……
编写一个程序来计算这两个序列不断变化的总和，直到达到某个次数。让用户交互地输入这个次数。
我自己编的程序如下所示：
#include <stdio.h>
int main(void)
{
    unsigned n, x, i;
    float sum1_n, sum2_n;
   
    while(1)
    {
        printf("请输入x的值：\n");
        i = scanf("%d", &x);
        if(i != 1)break;
        while(getchar() != '\n');
        sum1_n = 0;
        sum2_n = 0;
        
        if(x == 1)
        {
            sum1_n = 1;
            sum2_n = 1;
            printf("%f %f", sum1_n, sum2_n);
        }
        if(x == 2 * n)
        {
            for(n = 1; n <= x; n++)
            {
                sum2_n += 1 / (n-1) - 1 / n;
                sum1_n += 1 / n;
            }
            
            printf("%f %f", sum1_n, sum2_n);
        }
        else
        {
            for(n = 1; n <= x; n++)
            {
                sum2_n += 1 / n - 1 / (n - 1) + 1; // Thread1:EXC_ARITHMETIC(code=EXC_I386_DIV,subcode=0x0)[color=#FF0000]
                sum1_n += 1 / n;
            }
            printf("%f %f", sum1_n, sum2_n);
        }
        return 0;
    }
}

结果则如下所示
[local]1[/local]
输入任意结果都是显示错误，上方红字被提示有误，请示大神该如何修改呢？

n=1时除数为零，当然错了

#include <stdio.h>
int main(void)
{
    unsigned n, x, i;
    float sum1_n, sum2_n;
   
   
    while(1)
    {
        printf("请输入x的值：\n");
        i = scanf("%d", &x);
        if(i != 1)break;
        while(getchar() != '\n');
        sum1_n = 0;
        sum2_n = 0;
        
        if(x == 1)
        {
            sum1_n = 1;
            sum2_n = 1;
            printf("%f %f\n", sum1_n, sum2_n);
        }
        if(x % 2 == 0)
        {
            for(n = 2; n <= x; n++)
            {
                sum2_n += 1 / (n-1) - 1 / n;
                sum1_n += 1 / n;
            }
            
            printf("%f %f\n", sum1_n, sum2_n);
        }
        if((x % 2 == 1)&& (x != 1))
        {
            for(n = 2; n <= x; n++)
            {
                sum2_n += 1 / n - 1 / (n - 1) + 1;
                sum1_n += 1 / n;
            }
            printf("%f %f\n", sum1_n, sum2_n);
        }
        
    }
    return 0;
}
修改了下程序，然并软。

算法有问题
#include <stdio.h>
 int main(void)
 {
     unsigned n, x, i;
     float sum1_n, sum2_n,term2;
     
     
     while(1)
     {
         printf("请输入x的值：\n");
         i = scanf("%d", &x);
         if(i != 1&&getchar()=='q')break;
         while(getchar() != '\n');
         sum1_n=0,sum2_n=0;
         term2=-1;
         for(n=1;n<=x;n++)
         {
             sum1_n+=1.0/n;
             term2*=(-1.0)/n;
             sum2_n+=term2;
         }
         printf("%lf %lf\n",sum1_n,sum2_n);
     }
     return 0;
 }

[此贴子已经被作者于2016-8-19 21:42编辑过]

怎么算的－－我高数荒废了两年了哎

没有高数啥事吧。。。注意观察规律。。
当x=1    sum1=1/1    sum2=1/1
当x=2    sum1=1/1+1/2    sum2=1/1-1/2
当x=3    sum1=1/1+1/2+1/3    sum2=1/1-1/2+1/3
当x=n    sum1=1/1+....1/(n-1)+1/n    sum2=1/1-..+..-..+..?1/n(?号处是加是减反正是循环变换的，虽然和n%2==0有关系，但不必要提前考虑)

第一次：term2=1;第二次term2=-1/2，第三次term2=(-1/2)*(-1/3)=1/6

对不起，是错了
#include <stdio.h>
 int main(void)
 {
     unsigned n, x, i;
     float sum1_n, sum2_n,term2,sign;
     
     
     while(1)
     {
         printf("请输入x的值：\n");
         i = scanf("%d", &x);
         if(i != 1&&getchar()=='q')break;
         while(getchar() != '\n');
         sum1_n=0,sum2_n=0;
         term2=1;
         sign=1;
         for(n=1;n<=x;n++)
         {
             sum1_n+=1.0/n;
             sign*=-1.0;
             term2=sign*(-1.0)/n;
             sum2_n+=term2;
         }
         printf("%lf %lf\n",sum1_n,sum2_n);
     }
     return 0;
 }