Exercise

1.输出1000~2000之间的闰年

#include<stdio.h>int main(){int year;         printf("以下输出为闰年：\n");for(year=1000;year<=2000;year++){if(year%4==0 && year%100!=0 || year%400==0){printf("%d ",year);}}return 0;}

2.求1-1/2+1/3-1/4+.......+1/99-1/100的值。

#include<stdio.h>int main(){float sum=1;         float deno=2;float sign=1;float term;for(deno=2;deno<=100;deno++){sign=(-1)*sign;term=sign*(1/deno);sum=sum+term;}printf("sum=%f\n",sum);return 0;}