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description

有两个日期，求两个日期之间的天数，如果两个日期是连续的我们规定他们之间的天数为两天。

Input

有多组数据，每组数据有两行，分别表示两个日期，形式为YYYYMMDD

Output

每组数据输出一行，即日期差值

Sample Input Copy

20130101
20130105

Sample Output Copy

5

idea

• 闰年：2月29天，一年366天
  平年：2月28天，一年365天
  判断平年or闰年：
  能被4整除但不能被100整除 或者 能被400整除为闰年，
  否则为平年
• 注意日期相同为一天
  例如20220107-20220107相差一天（7号共一天
  20220107-20220108相差两天（7号和8号共两天

solution1

#include <cstdio>
#include <algorithm> 
using namespace std;
bool leap(int i){if(i % 400 == 0 || (i % 4 == 0 && i % 100 != 0))return true;elsereturn false;
}
int main(){int t[2], y[2], m[2], d[2], day, month[7] = {1, 3, 5, 7, 8, 10, 12}, temp, flag, flag1;while(scanf("%d%d", &t[0], &t[1]) != EOF){flag = 1;temp = 0;if(t[0] > t[1])swap(t[0], t[1]);for(int i = 0; i < 2; i++){y[i] = t[i] / 10000;m[i] = t[i] % 10000 / 100;d[i] = t[i] % 100;}day = (y[1] - y[0])*365;for(int i = y[0]; i < y[1]; i++)//闰年366天，多一天 if(leap(i))day++; if(m[0] > m[1]){flag = -1;swap(m[0], m[1]);swap(y[0], y[1]);}for(int i = m[0]; i < m[1]; i++){flag1 = 0;for(int j = 0; j < 7; j++){if(i == month[j]){temp += 31;flag1 = 1;break;}					}if(i == 2 && !flag1){if(leap(y[1]))temp += 29;elsetemp += 28;}else if(!flag1)temp += 30;}day += flag*temp + d[1] - d[0] + 1;printf("%d\n", day);}return 0;
}

solution2

#include <cstdio>
#include <algorithm> 
using namespace std;
int leap(int i){if(i % 400 == 0 || (i % 4 == 0 && i % 100 != 0))return 0;elsereturn 1;
}
int main(){int t[2], y[2], m[2], d[2], day, month[13][2] = {{0}, {31, 31}, {29, 28}, {31, 31}, {30, 30}, {31, 31}, {30, 30}, {31, 31}, {31, 31}, {30, 30}, {31, 31}, {30, 30}, {31, 31}};while(scanf("%d%d", &t[0], &t[1]) != EOF){int temp = 0, flag = 1;if(t[0] > t[1])swap(t[0], t[1]);for(int i = 0; i < 2; i++){y[i] = t[i] / 10000;m[i] = t[i] % 10000 / 100;d[i] = t[i] % 100;}day = (y[1] - y[0])*365;for(int i = y[0]; i < y[1]; i++)//闰年366天，多一天 if(!leap(i))day++; if(m[0] > m[1]){flag = -1;swap(m[0], m[1]);swap(y[0], y[1]);//需要相应年份判断平闰}while(m[0] < m[1]){temp += month[m[0]][leap(y[1])];m[0]++;}day += flag*temp + d[1] - d[0] + 1;printf("%d\n", day);}return 0;
}

solution3

#include <cstdio>
#include <algorithm>
using namespace std;
int leap(int n){if(n % 400 == 0 || (n % 4 == 0 && n % 100 != 0))return 1;elsereturn 0;
}
int main(){int l ,count = 1, month[12][2] = {{31, 31}, {28, 29}, {31, 31}, {30, 30}, {31, 31}, {30, 30}, {31, 31}, {31, 31}, {30, 30}, {31, 31}, {30, 30}, {31, 31}} ;int t[2], d[2] = {0}, m[2] = {0}, y[2] = {0};while(scanf("%d%d", t, t + 1) != EOF){count = 1;if(t[0] > t[1]) swap(t[0], t[1]);for(int i = 0; i < 2; i++){y[i] = t[i] / 10000;m[i] = t[i] % 10000 / 100;d[i] = t[i] % 100;}while(y[0] < y[1] || m[0] < m[1] || d[0] < d[1]){l = leap(y[0]);d[0]++;if(d[0] > month[m[0] - 1][l]){m[0]++;d[0] = 1;}if(m[0] > 12){y[0]++;m[0] = 1;}count++;}printf("%d\n", count);		}return 0;
}