//输入一个正整数，输出2000年1月1日经过该整数天后的日期.已测试，输入值可以为0~1095727//如，100天后，日期为2000 4 10#include<stdio.h>#define MAX_YEAR 5000//年数可以从2000一直到4999年。//函数功能：求解第year年共有多少天int day_in_year(int year){	if (((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0))		return 366;	else		return 365;}int main(){	int input;	int year, month, day;	long Accumulate_days_for_year[MAX_YEAR] = {0};	int Accumulate_days_for_month_in_leap_year[13] = {0,31,29,31,30,31,30,31,31,30,31,30,31};	int Accumulate_days_for_month_in_aver_year[13] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };	int i;	scanf_s("%d", &input);	for (i = 2000; i < MAX_YEAR; i++)	{		Accumulate_days_for_year[i] = day_in_year(i);	}	for (i = 2001; i < MAX_YEAR; i++)	{		//该数组[2000]为366；[2001]为366+365，即2000年和2001年天数的累和；[2002]为366+365+365，即2000-2002年天数的累和。依次类推。		//通过将累和与输入input比较，即可得知结果为哪一年。		Accumulate_days_for_year[i] = Accumulate_days_for_year[i - 1] + Accumulate_days_for_year[i];	}	for (i = 2; i < 13; i++)	{		//该数组[1]为31；[2]为31+29，即1月和2月天数的累和；[3]为31+29+31，即1-3月天数的累和。依次类推。		//通过累和判断结果为哪一月。		Accumulate_days_for_month_in_leap_year[i] = Accumulate_days_for_month_in_leap_year[i] + Accumulate_days_for_month_in_leap_year[i - 1];		//该数组[1]为31；[2]为31+28，即1月和2月天数的累和；[3]为31+28+31，即1-3月天数的累和。依次类推。		//通过累和判断结果为哪一月。		Accumulate_days_for_month_in_aver_year[i] = Accumulate_days_for_month_in_aver_year[i] + Accumulate_days_for_month_in_aver_year[i - 1];	}	//判断年数	for (i = 2000; i < MAX_YEAR; i++)	{		if (input < Accumulate_days_for_year[i])		{			year = i;			break;		}	}	//得知年数后，通过下式判断结果为当年的第几天。	input = input - Accumulate_days_for_year[year-1];	//如果year为闰年，使用月累和数组Accumulate_days_for_month_in_leap_year	if (((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0))	{		for (i = 1; i < 13; i++)		{			if (input < Accumulate_days_for_month_in_leap_year[i])			{				month = i;				day = input - Accumulate_days_for_month_in_leap_year[i - 1] + 1;				break;			}		}	}	//如果year为平年，使用月累和数组Accumulate_days_for_month_in_aver_year	else 	{		for (i = 1; i < 13; i++)		{			if (input < Accumulate_days_for_month_in_aver_year[i])			{				month = i;				day = input - Accumulate_days_for_month_in_aver_year[i - 1] + 1;				break;			}		}	}	printf("%d\n", year);	printf("%d\n", month);	printf("%d\n", day);	return 0;}