c语言;输入一个年份,输出该年各月份天数以及总天数。

在线急等

#include <stdio.h>
main()
{
int i,year,month[12] ={31,28,31,30,31,30,31,31,30,31,30,31};
scanf("%d",&year);
if((year%4==0)&&(year%100!=0)||(year%400==0))month[1]++;year=0;
for (i=0;i<12;i++)
printf("month[%d]:day[%d]\n",i+1,month[i]),year+=month[i];
printf("year:[%d]days\n",year);
}


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第1个回答  2013-05-03
#include <stdio.h>
#include<conio.h>
int day_of_year(int year);
static char pdata[][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 main()
{
int n;
int i;
int year;
printf("请输入年份:");
scanf("%d",&year);

n=day_of_year(year);
if(n==-1)
printf("month_day failed\n");
else if(n==1)
{
printf("今年是平年,共366天\n");
for(i=0;i<13;i++)
printf("%2d",pdata[1][i]);
}
else
{
printf("今年是闰年,共365天\n");
for(i=0;i<13;i++)
printf("%4d",pdata[0][i]);
}
return 0;
}
int day_of_year(int year)
{
int i;
int pdays=0;
int leap;
if(year<1752)
return -1;
if(year%4==0&&year%100!=0||year%400==0)
leap=1;
else
leap=0;
return leap;
}
第2个回答  2013-05-03
#include <stdio.h>
int main()
{
int year,month,days,t;
printf("Please input year\n");
scanf("%d",&year);
for(month=1;month<=12;month++){
if(month==1||month==3||month==5||month==7||month==8||month==10||month==12)
days=31;
else if(month==2){
if((year%4==0)&&(year%100!=0)||(year%400==0)){
days=29;
t=366;
}

else{
days=28;
t=365;
}

}
else
days=30;
printf("month%d:days=%d\n",month,days);
}
printf("总天数:%d\n",t);
}
第3个回答  2013-05-03
void OutPut(UINT nYear)
{
BYTE mon[12] = {31,28,31,30,31,30,31,31,30,31,30,31};
UNIT nSum = 365;
if(y%400==0||(y%4==0&&y%100!=0))
{
mon[1]=29;
nSum = 366;
}
for(int i = 0 ; i < 12 ; i++)
{
printf("mon%d %d\n",i + 1 , mon[i]);
}
printf("total %d\n",nSum);

}本回答被提问者采纳
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