int one() { int n,m; n = 4; m = 6; while(n > 0) { n = n - 1; m = m + 2; } return m; }
int f(int x) { if(x < 2) { return x + 1; } else { return 2*x + 1; } }
void r(int a, int x, int &z) { cout << r: a = " << a << " x = " << x << " z = " << z << endl; a = 2; x = 5; z++; } void t() { int a, b, x, z; a = 9; b = 20; x = 67; z = 92; r(a,b,x); r(a,b,x); cout << "t: a = " << a << " b = " << b << " x = " << x << " z = " << z << endl; }
int mystery(int n) { if(n == 0) return 0; else return mystery(n-1) + n; }
Note that you can get the rightmost digit of a number n by taking the remainder when you divide n by 10, and you can remove the rightmost digit by dividing n by 10.
For this problem, do not use any form of loop, and do not alter the value of any variable that already has a meaningful value.
int sevens(long n)