//计算 1/1-1/2+1/3-1/4+1/5......+1/99-1/100的值,打印结果 //法一: int main() { int i = 0; double sum = 0.0;//1.0时,用double sum int flag = 1; for (i = 1; i <= 100; i++) { sum += flag * 1.0 / i;//1/i时,1/1=,1/2=0...(错误),所以用1.0 flag = -flag; //使数字在1 -1 持续循环发生变化 } printf("%lf", sum);//double,用%lf return 0; } //法二: int main() { int i = 0; double sum = 0.0;//1.0时,用double sum for (i = 1; i <= 100; i++) { if (i % 2 == 0) sum -= 1.0 / i; else sum += 1.0 / i; } printf("%lf", sum);//double,用%lf return 0; }