【程序4】 题目：输入某年某月某日，判断这一天是这一年的第几天？

方法一：

直接模拟，首先累加平年的天数，如果是闰年且月份数大于2，那么就加1。

#include <stdio.h>

#define judge(x) ((x%4==0 && x%100!=0)||(x%400==0))

int main(){

int year,month,day,count;

while(~scanf("%d%d%d",&year,&month,&day)){

count = 0;

switch(month){

case 1:count += 0; break;
case 2:count += 31; break;
case 3:count += 59; break;
case 4:count += 90; break;
case 5:count += 120; break;
case 6:count += 151; break;
case 7:count += 181; break;
case 8:count += 212; break;
case 9:count += 243; break;
case 10:count += 273; break;
case 11:count += 304; break;
case 12:count += 334; break;
}

count += (month>2 && judge(year)) + day;

printf("%d\n",count);
}

return 0;
}


方法二

应该更容易实现一些，使用数组存储天数，然后累加出结果。

#include <stdio.h>

const int monthA[13] = {0,31,28,31,30,31,30,31,31,30,31,30,31};
const int monthB[13] = {0,31,29,31,30,31,30,31,31,30,31,30,31};

int main(){

int year,month,day;
const int *p = NULL;
int i,count;

while(~scanf("%d%d%d",&year,&month,&day)){

count = 0;

if((year%4==0 && year%100!=0)||(year%400==0))
p = monthB+1;

else
p = monthA+1;

for(i=1;i<=month-1;++i,++p)
count += *p;

printf("%d\n",count+day);
}

return 0;
}


方法三

把方法一的顺序颠倒一下，省去人力求和的过程。

#include <stdio.h>

#define judge(x) ((x%4==0 && x%100!=0)||(x%400==0))

int main(){

int year,month,day;
int count;

while(~scanf("%d%d%d",&year,&month,&day)){

count = 0;

switch(month){

case 12:count += 30;
case 11:count += 31;
case 10:count += 30;
case 9:count += 31;
case 8:count += 31;
case 7:count += 30;
case 6:count += 31;
case 5:count += 30;
case 4:count += 31;
case 3:count += 28 + judge(year);
case 2:count += 31;
case 1:count += 0;
}

count += day;

printf("%d\n",count);
}

return 0;
}