版权声明:关中大侠Lv轻侯 https://blog.csdn.net/weixin_44312186/article/details/88064599
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。
输入
有多组数据,每组数据有两行,分别表示两个日期,形式为YYYYMMDD
输出
每组数据输出一行,即日期差值
样例输入
20130101 20130105
样例输出
5
代码一(codeup超时) :
#include<stdio.h>
#include <stdbool.h>
int mouth[13][2]={{0,0},
{31,31},{28,29},{31,31},{30,30},{31,31},{30,30},
{31,31},{31,31},{30,30},{31,31},{30,30},{31,31},
};
bool isleap(int year){ //判断是否为闰年
return (year%4==0&&year%100!=0)||(year%400==0);
}
int main(){
int time1,y1,m1,d1;
int time2,y2,m2,d2;
while(scanf("%d%d",&time1,&time2)!=EOF){
if(time1>time2){ //第一个日期晚于第二个日期则交换
int temp=time1;
time1=time2;
time2=temp;
}
y1=time1/10000,m1=time1%10000/100,d1=time1%100;
y2=time2/10000,m2=time2%10000/100,d2=time2%100;
int ans=1;//记录结果
while(y1<y2||m1<m2||d1<d2){
d1++;//天数加一
if(d1==mouth[m1][isleap(y1)]){ //已够当月天数
m1++;
d1=1;
}
if(m1==13){
y1++;
m1=1;
}
ans++;//累计
}
printf("%d\n",ans);
}
return 0;
}
ps:超时主要是由于if(d1==mouth[m1][isleap(y1)])判断错误导致。
正确代码 :
#include<stdio.h>
#include <stdbool.h>
int mouth[13][2]={{0,0},
{31,31},{28,29},{31,31},{30,30},{31,31},{30,30},
{31,31},{31,31},{30,30},{31,31},{30,30},{31,31},
};
bool isleap(int year){ //判断是否为闰年
return (year%4==0&&year%100!=0)||(year%400==0);
}
int main(){
int time1,y1,m1,d1;
int time2,y2,m2,d2;
while(scanf("%d%d",&time1,&time2)!=EOF){
if(time1>time2){ //第一个日期晚于第二个日期则交换
int temp=time1;
time1=time2;
time2=temp;
}
y1=time1/10000,m1=time1%10000/100,d1=time1%100;
y2=time2/10000,m2=time2%10000/100,d2=time2%100;
int ans=1;//记录结果
while(y1<y2||m1<m2||d1<d2){
d1++;//天数加一
if(d1==mouth[m1][isleap(y1)]+1){ //已够当月天数
m1++;
d1=1;
}
if(m1==13){
y1++;
m1=1;
}
ans++;//累计
}
printf("%d\n",ans);
}
return 0;
}