日期差值

计算两个日期相对于0000 00 00的差值
然后差值相减取绝对值+1就可以了。

#include<iostream>
using namespace std;
int daytab[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}
};
bool isLeapYear(int year){
    if(year%100!=0&&year%4==0||year%400==0)
        return true;
    return false;
}
char a[9],b[9];
int main(){
    //n为第一个日期相对于0000 00 00的差值，n1为为第二个的差值
    int year,month,day,year1,month1,day1,n,n1;
    while(cin>>a>>b){
        n=n1=0;
        sscanf(a,"%4d%2d%2d",&year,&month,&day);//格式化读入
        sscanf(b,"%4d%2d%2d",&year1,&month1,&day1);
        for(int i=0;i<=year;i++){//记录年差值
            if(isLeapYear(i)){
                n+=366;
            }else n+=365;
        }for(int i=1;i<month;i++){//记录月差值
            n+=daytab[isLeapYear(year)][i];
        }n+=day;//记录日差值
        for(int i=0;i<=year1;i++){
            if(isLeapYear(i)){
                n1+=366;
            }else n1+=365;
        }for(int i=1;i<month1;i++){
            n1+=daytab[isLeapYear(year1)][i];
        }n1+=day1;
        cout<<abs(n-n1)+1<<endl;//差值相减取绝对值+1即可
    }
    return 0;
}