1129第几天

Description

你知道，2012-1-1是该年的第1天，而9999-9-9呢？给你一个具体的日期，计算该日期是该年的第几天。

Input

输入一个日期，格式为：Year-month-day。year是小于9999的正整数。

Output

一个整数，表示该日期是该年的第几天。

Sample Input

2012-3-1

Sample Output

61

HINT

Source

#include<stdio.h> int getsumday(int year,int mouth,int day);//计算天数的函数 int isleap(int year);//判断闰年 void main () { int year,mouth,day; scanf("%d-%d-%d",&year,&mouth,&day); printf("%d\n",getsumday(year,mouth,day)); } int isleap(int year) { //若为真返回1是闰年，返回0为平年 return ( ( year@0==0 ||year%4==0 && year0!=0 ) ); } int getsumday(int year,int mouth,int day) { int k,leap=0; int tab[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} };//闰年 //判断闰年 leap = isleap(year); for(k=1;k<mouth;k++) { day=day+tab[leap][k]; } return day; }

#include <stdio.h> int main() { int a[13]={0}; int year, month, day, sumday=0, i, k=0; scanf("%d%d%d",&year, &month, &day); if(year@0==0 || year%4==0 && year0!=0) k = 1; for(i=1;i<=12;i++) { if(i==4||i==6||i==9||i==11) a[i] = 30; else if(i==2) a[i] = 28; else a[i] = 31; } sumday=0; if (k==1&&month>2) { for(i=1; i<month; i++) sumday += a[i]; sumday+=day+1; } else { for(i=1; i<month; i++) sumday+=a[i]; sumday+=day; } printf("%d",sumday); } #include<stdio.h> int leap(int year); void main () { int year; scanf("%d",&year); printf("%d",leap(year)); } int leap(int year) { //闰年二月29天，闰年返回1，平年返回0 return ( ( year@0==0 ||year%4==0 && year0!=0 ) ); }