int m = 0, n= 0, k = 0, sum = 0; m = a/100; sum = sum + m; n = (a - m * 100)/10; sum = sum + n; k = a - m * 100 - n * 10; sum = sum + k; printf ("这个三位整数各位数之和是:%d", sum); printf ("\n");
}
void count(int a[4])
{ int i = 0,m = 0, n = 0; for (;i < 4;i++) { if (0 == (a[i] % 2)) m = m + 1; else n = n + 1; } printf ("偶数有%d个\n",m); printf ("奇数有%d个\n",n);
}
int main(void)
{ int i = 0, n, a[4]; printf ("请输入一个三位整数\n"); scanf ("%d",&n); add(n); printf ("请输入整形数组数据\n"); for (;i < 4;i++) { scanf ("%d",&a[i]); getchar(); } count(a[4]); return 0;
}
时间紧迫最后一个就不写了.这两个还是可以运行的.