1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 using namespace std;
 5 
 6 int tianxia[2][12]={
 7     {31,28,31,30,31,30,31,31
 
,30,31,30,31},
 8     {31,29,31,30,31,30,31,31,30,31,30,31}
 9 };
10 
11 int isPrime(int year){
12     if(year%4==0&&year%100!=0 || year%400==0)
13         return 1;
14     else
15         return 0;
16 }
17 
18 int main()
19 {
20     int a,b;
21     scanf("%d%d",&a,&b);
22     if(a<b)
23         a^=b;b^=a;a^=b;
 
24     ///經過上邊的操作後，日期a將大於日期b
25     int yeara=a/10000;
26     int moutha=(a%10000)/100;
27     int daya=(a%100);
28 
29     int yearb=b/10000;
30     int mouthb=(b%10000)/100;
31     int dayb=(b%100);
32 
33     int day1,day2,day3;
34     day1=day2=day3=0;
35 
36     for(int i=0;i<moutha-1;i++){
37             printf("%d : %d\n",i+1,tianxia[isPrime(yeara)][i]);
38         day1+=tianxia[isPrime(yeara)][i];
39     }
40     day1+=daya-1;
41 
42     for(int i=0;i<mouthb-1;i++){
43             printf("%d : %d\n",i+1,tianxia[isPrime(yearb)][i]);
44         day2+=tianxia[isPrime(yearb)][i];
45     }
46     day2+=dayb-1;
47 
48     for(int i=yearb;i<yeara;i++){
49         if(isPrime(i)==0)day3+=365;
50         else day3+=366;
51     }
52     int ans = day3+day1-day2;
53     printf("The ans is :%d  %d  %d  %d",ans,day1,day2,day3);
54     return 0;
55 }

如何計算兩個（年-月-日）中間的間隔天數