poj3393(模拟）

题意：好长。。。给出一个时间段，问在这个时间段中，每个月第一个工作日为星期一（goodday）和最后一个工作日为星期五（luckyday）的数目各为多少。

思路：判断闰月后，不断加就好了，具体看程序。

#include<iostream>

using namespace std;
#define DAY ((day+5)%7)

int test, day, ys, ms, ye, me, good, lucky;
int month[12]={31,28,31,30,31,30,31,31,30,31,30,31};

bool leap(int i){
	if (i < 1582){
		if (i % 4 == 0) return true;
	}
	else{
		if (i == 1700){
			return true;
		}
		if (((i % 4 == 0) && (i % 100 != 0)) || (i % 400 == 0))
			return true;
	}
	return false;
}

int to_today(int ys, int ms){
	for (int i = 1; i < ys; i++){
		if (leap(i)) day+=366;
		else day+= 365;
		day %= 7;
	}

	if (ys > 1752) day = (day + 3)%7;
	for (int i = 1; i < ms; i++){
		day += month[i-1];
		if (leap(ys) && i == 2) day+=1;
		else if (ys == 1752 && i == 9) day-=11;
	}
	day %= 7;

	return day;
}

int main(){

	scanf("%d", &test);
	while(test--){
		day = 1;
		good = lucky = 0;
		scanf("%d%d%d%d",&ys,&ms,&ye,&me);
		day = to_today(ys, ms) % 7;

		int year = ys;
		for (int i = ms; i <= 12;){
			//cout << "firstDay: " <<DAY << endl;
			if (DAY < 2 || DAY == 6) good++;
			day += month[i-1];
			if (leap(year) && i == 2) day++;
			if (year == 1752 && i == 9) day -= 11;
			day %= 7;
			if (DAY == 6 || DAY < 2) lucky++;
			//cout << "lastDay: " << DAY << endl;
			if (i==me && year==ye) break;
			i++;
			if (i == 13){
				i = 1;
				year++;
			}
		}

		printf("%d %d\n", lucky, good);
	}
}