给出年、月、日，计算该日时该年的第n天

给出年、月、日，计算该日时该年的第n天
急

用C语言，不要c++

#include
void main()
{int Y,M,D,a[13]={366,31,29,31,30,31,30,31,31,30,31,30,31},
b[13]={365,31,28,31,30,31,30,31,31,30,31,30,31},E,i;char c;

K:scanf("%d%c%d%c%d",&Y,&c,&M,&c,&D);
if((Y%4==0&&Y%100!=0)||(Y%4==0&&Y%400==0))
{E=0;
for(i=1;i E=E+a[i];
printf("It is the year's first %d day.\n",E+D);}
else
{E=0;
for(i=1;i E=E+b[i];
printf("It is the year's first %d day.\n",E+D);}
goto K;
getch();}
注:将注释删除之后再运行，输入格式形如，2009,12,25或2009/12/25或2009-12-25或
2009 12 25或2009.12.25或2009\12\25或2009.12/25等都是可以的，还可以是其他的形式，不过注意在数字之间必须插入一个符号就是了

#include #include int main() { int Y,M,D; scanf("%d%d%d",&Y,&M,&D); int i,n,n1=0; int m[12]={31,28,31,30,31,30,31,31,30,31,30,31}; if((Y%4==0&&Y%100!=0)||Y%400==0) m[1]=29; for(i=2;i<=M;i++) n1+=m[i-2]; n=n1+D; printf("%d",n); return 0; } 对于任意给定的年、月、日都行。

void main() { int year=2009,month=12,day=25; bool leap=0; leap = (year%400==0)| (year%100!=0) & (year%4==0);//是否为闰年 for (i=1;i<=month-1;i++) //当年的累计天数 switch (i) { case 1: case 3: case 5: case 7: case 8: case 10: case 12: total+=31; break; case 4: case 6: case 9: case 11: total+=30; break; case 2: if ...void main() { int year=2009,month=12,day=25; bool leap=0; leap = (year%400==0)| (year%100!=0) & (year%4==0);//是否为闰年 for (i=1;i<=month-1;i++) //当年的累计天数 switch (i) { case 1: case 3: case 5: case 7: case 8: case 10: case 12: total+=31; break; case 4: case 6: case 9: case 11: total+=30; break; case 2: if (leap) total+=29; else total+=28; break; } total = total+day; //加上当月的天数 }

等下给你