编程求1-1/2+1/3-1/4+…+1/99-1/100

以下是两种编程求解1-1/2+1/3-1/4+…+1/99-1/100的例子：

1. 使用C语言编程求解：

#include <stdio.h>

int main() {
    int i;
    double sum = 0;
    int flag = 1;
    
    for (i = 1; i <= 100; i++) {
        sum = sum + flag * (1.0 / i);
        flag = -flag;
    }
    
    printf("1-1/2+1/3-1/4+...+1/99-1/100的值为：%lf", sum);
    
    return 0;
}

2. 使用C语言编程求解，先求1+1/3+1/5+...+1/99，再求1/2+1/4+...+1/100，后结果相减：

#include <stdio.h>

int main() {
    float i, j;
    float sum = 0, sum1 = 0, sum2 = 0;
    
    for (i = 1; i <= 100; i += 2) {
        sum1 = sum1 + 1 / i;
    }
    
    printf("总和sum1=%f\n", sum1);
    
    for (j = 2; j <= 100; j += 2) {
        sum2 = sum2 + 1 / j;
    }
    
    printf("总和sum2=%f\n", sum2);
    
    sum = sum1 - sum2;
    
    printf("sum=%.2f\n", sum);
    
    return 0;
}