int leapdays[13]={0,31,29,31,30,31,30,31,31,30,31,30,31}; in

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int leapdays[13]={0,31,29,31,30,31,30,31,31,30,31,30,31}; int a[13]; int sum,total; sum=0; a[0]=0;s
int leapdays[13]={0,31,29,31,30,31,30,31,31,30,31,30,31};
int a[13];
int sum,total;
sum=0;
a[0]=0;sum=0;
for(int i=1;i

int leapdays[13]={0,31,29,31,30,31,30,31,31,30,31,30,31}; //保存每月有多少天
int a[13]; //保存从1月1日到几月份有多少天
int sum,total; // sum 保存闰年数,total 保存周几
sum=0; a[0]=0;sum=0;
for(int i=1;i

int leapdays[13]={0,31,29,31,30,31,30,31,31,30,31,30,31}; in fun(int x,int y,int z) {z=x*x+y*y;} main() {int a=31; fun(5, 计算机位操作：int abs( int x ) { int y ; y = x > > 31 ; return (x^y main(){int a[]={0,31,28,31,30,31,30,31,31,30,31,30,31};/*平年的 #include void main(){int a[30];int i;for (i=0,i 31、下列程序段中可实现1×2×3×4×5×6×7×8×9×10的是（）.A、int k=1,m=0; for(;k int a[]={5,10,15,20,25,30}; int b=LA(a,4); int c=LA(a+2,3);p hour=hour*30+int(Minute/2)是什么意思 int a=2; int f(int a); {return (a)++;} main() {int s=0; {int #includeint x1=30,x2=40;void sub(int x,int y){x1=x;x=y;y=x1; 31．设有下列变量说明：char a; int x; float y; double z; 则表达式a*x-z+y的值的 30)有以下程序 #include main() { int b[3] [3]={0,1,2,0,1,2,0,1,2},