日期计算

//zzuli 1878
1.

#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
", 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
", sum);
        }
    }
    return 0;
}
2.另解：我直接采用了1年1月1日到两个输入的日期分别有多少天然后计算差值。这样的话关于闰年比较好计算，可以用容斥...闰年数量为x/4-x/100+x/400...

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<ctype.h>
#include<algorithm>
using namespace std;
#define N 100100
#define INF 0x3f3f3f3f

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

int js(int y, int m, int d)
{
    int sum=0, t, leap;
    t=(y-1)/4-(y-1)/100+(y-1)/400;
    sum+=365*(y-1)+t;
    leap=IsLeap(y);
    for(int i=1; i<m; i++)
        sum+=tab[leap][i];
    sum+=d;
    return sum;
}

int main()
{
    int T, y1, m1, d1, y2, m2, d2, ans1, ans2;
    scanf("%d", &T);

    while(T--)
    {
        scanf("%d%d%d%d%d%d", &y1, &m1, &d1, &y2, &m2, &d2);
        ans1=js(y1, m1, d1);
        ans2=js(y2, m2, d2);
        printf("%d
", ans2-ans1);
    }
    return 0;
}