#includeusing std::cin;using std::cout;using std::endl;void main(void){ int i, leap, month; int y1, y2, m1, m2, d1, d2; long day; int m[] = {31,28,31,30,31,30,31,31,30,31,30,31}; y1 = y2 = 0;m1 = m2 = 0; d1 = d2 = 0; leap = 0; day = 0; while((y1 12)||(d1m[m1-1])) { cout>y1>>m1>>d1; if((y1%4 == 0)&&(y1%25 != 0)||(y1%400 == 0)) m[1] = 29;else
m[1] = 28;
cout22)||(d2m[m2-1])) { cout>y2>>m2>>d2; if((y2%4 == 0)&&(y2%25 != 0)||(y2%400 == 0)) m[1] = 29;else
m[1] = 28;
cout<<endl; }for(i = y1; i < y2; i++) {if((i%4 == 0)&&(i%25 != 0)||(i%400 == 0)) leap++;day += 365; } day += leap; for(i = 0, month = d1; i < m1-1; i++) {month += m; } day -= month; for(i = 0, month = d2; i < m2-1; i++) {month += m; } day += month; cout<<y1<<" "<<m1<<" "<<d1<<"到" <<y2<<" "<<m2<<" "<<d2<<"的天数为"<<day<<endl;} 很简单的思路先计算中间有多少个整年，然后再计算出生年月日是那一年的第多少天，结束年月日又是那一年的第多少天，减一下，再加上整年的天数，就是所求代码长度是2L的1/4～～ |
