1503 - 日期差值
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
Examples
Input
20130101 20130105
Output
5
Solution C
#include <stdio.h> #include <math.h> int m[][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}; int leap(int year) { return((year%4==0&&year%100!=0)||(year%400==0)); } int day(int year,int month,int day) { int d=0; int k=0; int i; if(leap(year)) k=1; for(i=1;i<month;i++) d+=m[i][k]; d+=day; return(d); } main() { int y1,y2,m1,m2,d1,d2,day1,day2,i; while(scanf("%4d%2d%2d",&y1,&m1,&d1)!=EOF) { scanf("%4d%2d%2d",&y2,&m2,&d2); day1=day(y1,m1,d1); day2=day(y2,m2,d2); for(i=0;i<y1;i++) { if(leap(i)) day1+=366; else day1+=365; } for(i=0;i<y2;i++) { if(leap(i)) day2+=366; else day2+=365; } printf("%d\n",(int)fabs(day1-day2)+1); } }
Solution C++
#include<stdio.h> #include<math.h> int m[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}; int leap(int y) { return (y%4==0&&y%100!=0)||(y%400==0); } int day(int Y,int M,int D) { int k=0,i,d; if(leap(Y)) k=1; for(d=0,i=1;i<M;i++) d+=m[i][k]; d+=D; return d; } int main() { int Y1,M1,D1,Y2,M2,D2,i,d1,d2; while(scanf("%4d%2d%2d",&Y1,&M1,&D1)!=EOF) { scanf("%4d%2d%2d",&Y2,&M2,&D2); d1=day(Y1,M1,D1); d2=day(Y2,M2,D2); for(i=0;i<Y1;i++) if(leap(i)) d1+=366; else d1+=365; for(i=0;i<Y2;i++) if(leap(i)) d2+=366; else d2+=365; printf("%d\n",(int)fabs(d1-d2)+1); } return 0; }