zzuli 1878 计算两日期距离天数

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1878: 蛤玮准备礼物

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 595 Solved: 75

SubmitStatusWeb Board

Description

蛤玮的妹子要过生日了,经过千挑万选,蛤玮决定给妹子买一盒水彩铅笔,然而卖多少个才合适呢.机智的蛤玮决定数一数今天是妹子出生的第多少天,然后就买那么多根铅笔.

给出蛤玮妹子的生日和今天的日期,求蛤玮一共需要买多少.

注意:生日当天算第0天.

Input

T(1<=T<=1000),表示数据组数.

每组数据一行y1,m1,d1,y2,m2,d2,其中y1,m1,d1表示妹子生日的年月日,y2,m2,d2表示蛤玮所在时空的当前天的年月日.

(1900<=y<=1000000, 1<=m<=12, 1<=d<=当月应有天数).输入保证当前天在妹子生日之后.

Output

每组数据输出一行,蛤玮要买多少根铅笔.

Sample Input

21995 10 24 1995 10 251996 1 9 1996 3 7

Sample Output

HINT

Source

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<ctype.h>
#include<algorithm>
using namespace std;
#define N 1010000
#define INF 0x3f3f3f3f

int yearday[N];

int IsLeap(int year)
{
    return ((year%4==0&&year%100!=0) || year%400==0);
}

int main()
{
    int T, y1, m1, d1, y2, m2, d2, sum;
    int tab[2][13]={{0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},
    {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}};
    int leap;
    yearday[1899]=0;
    for(int i=1900; i<=1000000; i++)
        yearday[i]=yearday[i-1]+(IsLeap(i)==1 ? 366 : 365);
    scanf("%d", &T);

    while(T--)
    {
        scanf("%d%d%d%d%d%d", &y1, &m1, &d1, &y2, &m2, &d2);
        sum=0;
        if(y1==y2)
        {
            leap=IsLeap(y1);
            for(int i=m1+1; i<m2; i++)
                sum+=tab[leap][i];
            if(m1<m2)
            {
                sum+=tab[leap][m1]-d1;
                sum+=d2;
            }
            else
                sum+=d2-d1;
            printf("%d\n", sum);
        }
        else
        {
            leap=IsLeap(y1);
            for(int i=m1+1; i<=12; i++)
                sum+=tab[leap][i];
            sum+=tab[leap][m1]-d1;
            leap=IsLeap(y2);
            for(int i=1; i<m2; i++)
                sum+=tab[leap][i];
            sum+=d2;
            sum+=yearday[y2-1]-yearday[y1];
            printf("%d\n", sum);
        }
    }
    return 0;
}

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