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题目描述

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

输出描述:

示例1

20110412

输出

11

代码

#include 
   
    #include 
    
     #include 
     
      #include 
      
       using namespace std;int datas[2][12]={
  {31,28,31,30,31,30,31,31,30,31,30,31},                  {31,29,31,30,31,30,31,31,30,31,30,31}};int isLeapYear(int year){    return (year%4==0&&year%100!=0)||year%400==0;}int numberOfYear(int year){    if(isLeapYear(year))         return 366;    else         return 365;}// 思路就是 年计算相对值，日期计算本年的值，然后多的年日期-少的年日期；+年的日子；int main(){    string time1;    string time2;    int year,month,day,year2,month2,day2;    while(getline(cin,time1)&&getline(cin,time2)){        int num=1; //更好的读入方法：        int flag = time1.compare(time2);        year=stoi(time1.substr(0,4));        year2=stoi(time2.substr(0,4));        month=stoi(time1.substr(4,2));        month2=stoi(time2.substr(4,2));        day=stoi(time1.substr(6,2));        day2=stoi(time2.substr(6,2));                int numOfYear=abs(year-year2);        // 函数化处理：        int num1,num2;        num1=num2=0;        for(int i=0;i
       
        0){// first is biger            while(numOfYear>0){                num+=numberOfYear(year);                year--;                numOfYear--;            }            num+=(num1-num2);        }else{             while(numOfYear>0){                num+=numberOfYear(year);                year++;                numOfYear--;            }            num+=(num2-num1);        }        printf("%d\n",num);    }        return 0;}
       
      
     
    
   

优秀答案：

#include
   
    #define leap(x) (x%400==0||(x%4==0&&x%100!=0)?1:0)#define Abs(x) ((x)>0?(x):-(x))typedef struct date{    int y, m, d;}Date;const int yd[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 days(Date x){    int sum=0, i;    int y=x.y, m=x.m, d=x.d;    for(i=0; i
   

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