第9章练习题参考解答.doc

第9章 函数2. was #2 int check(int x, int y, int n) return (x = 0 & x = 0 & y = n - 1);3.int gcd(int m, int n) int r; while (n != 0) r = m % n; m = n; n = r; return m;4. was #4 int day_of_year(int month, int day, int year) int num_days = 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31; int day_count = 0, i; for (i = 1; i 2) day_count+; return day_count + day;5.int num_digits(int n) int digits = 0; do n /= 10; digits+; while (n != 0); return digits;6. was #6; modified int digit(int n, int k) int i; for (i = 1; i k; i+) n /= 10; return n % 10;10. was #10 (a)int largest(int a, int n) int i, max = a0; for (i = 1; i max) max = ai; return max;(b)int average(int a, int n) int i, avg = 0; for (i = 0; i n; i+) avg += ai; return avg / n;(c)int num_positive(int a, int n) int i, count = 0; for (i = 0; i 0) count+; return count;11.float compute_GPA(char grades, int n) int i, total = 0; for (i = 0; i n; i+) switch (toupper(gradesi) case A: total += 4; break; case B: total += 3; break; case C: total += 2; break; case D: total += 1; break; return (float) total / n;12.double inner_product(double a, double b, int n) int i; double total = 0.0; for (i = 0; i n; i+) total += ai * bi; return total;13.int evaluate_position(char board88) int i, j, value = 0; for (i = 0; i 8; i+) for (j = 0; j 8; j+) switch (boardij) case Q: value += 9; break; case R: value += 5; break; case B: case N: value += 3; break; case P: value += 1; break; case q: value -= 9; break; case r: value -= 5; break; case b: case n: value -= 3; break; case p: value -= 1; break; return value;14. #define true 1#defne false 0#define unknown -1int has_zero(int a, int n) int i;flag1=0,flag2=n; for (i = 0; i n; i+) if (ai = 0) flag1+;else flag2-;if(flag1=n) return true;else if(flag2=0) return false;else return unknown;15. was #12; modified double median(double x, double y, double z) double result; if (x = y) if (y = z) result = y; else if (x = z) result = z; else result = x; else if (z = y) result = y; else if (x = z) result = x; else result = z; return result;16.int fact(int n) return n = 1 ? 1: n * fact(n - 1);17. was #14 int fact(int n) int i, result = 1; for (i = 2; i = n; i+) result *= i; return result;18.int gcd(int m, int n) return n = 0 ? m : gcd(n, m % n);19. was #16 使用下列代码测试 pb函数: #include void pb(int n);int main(void) int n; printf(Enter a number: ); scanf(%d, &n); printf(Output of pb: ); pb(n); printf(n); return 0;void pb(int n) if (n != 0) pb(n / 2); putchar(0 + n % 2); 实际上，pb的作用就是打印出n对应的二进制数（用字符方式，假设n0）。例如： Enter a number: 53Output of pb: 110101A trace of pbs execution would look like this: pb(53) finds that 53 is not equal to 0, so it calls pb(26), which finds that 26 is not equal to 0, so it calls pb(13), which finds that 13 is not equal to 0, so it calls pb(6), which finds that 6 is not equal to 0, so it calls pb(3), which finds that 3 is not equal to 0, so it calls pb(1), which finds that 1 is not equal to 0, so it calls pb(0), which finds that 0 is equal to 0, so it returns, causingpb(1) to print 1 and return, causingpb(3) to print 1