请在 下方输入 要搜索的题目：
推荐参考答案 (
由 搜题小帮手 官方老师解答 )
答案：
以下文字与答案无关
提示：有些试题内容 显示不完整,文字错误 或者 答案显示错误等问题，这是由于我们在扫描录入过程中 机器识别错误导致,人工逐条矫正总有遗漏，所以恳请 广大网友理解。
相关试题
- n) m = m - n; while (n > m) n = n - m; } printf("m=%d",m); return 0;}选项： A:m=3; B:m=2; C:m=1; D:m=0">1.运行以下程序后,如果从键盘上输入65 14<回车>,则输出结果为()。 int main(void){ int m, n; printf("Enter m,n;"); scanf("%d%d", &m,&n); while (m != n) { while (m > n) m = m - n; while (n > m) n = n - m; } printf("m=%d",m); return 0;}选项： A:m=3; B:m=2; C:m=1; D:m=0
- 2.运行以下程序后，输出结果为（）。#include int main(){ int m=65, n=14; while (m != n) { while (m > n) m = m - n; while (n > m) n = n - m; } printf("m=%d",m); return 0;}选项： A:m=3 B:m=2 C:m=1 D:m=0
-
3.以下程序是求整数m,n的最大公约数和最小公倍数，请补充完整： int gys(m,n) int m,n; { int t,r;if(m
- = 0) { if (m%2 == 0) printf("%d is a positive even\n", m); else printf("%d is a positive odd\n", m); } else { if (m % 2 == 0) printf("%d is a negative even\n", m); else printf("%d is a negative odd\n", m); } } 若键入- 9 ,则运行结果为 :">4.#include main() { int m; scanf("%d", &m); if (m >= 0) { if (m%2 == 0) printf("%d is a positive even\n", m); else printf("%d is a positive odd\n", m); } else { if (m % 2 == 0) printf("%d is a negative even\n", m); else printf("%d is a negative odd\n", m); } } 若键入- 9 ,则运行结果为 :
-
5.【填空题】以下程序是求整数m,n的最大公约数和最小公倍数，请补充完整： int gys(m,n) int m,n; { int t,r;if(m
- 6.完成一下程序求出两个正整数的最大公约数。#include "stdio.h"void main(){ int r,m,n; printf("输入m,n:"); scanf("%d %d",m,n); r=m%n; while(______ ) { m=______ ; n=r; r=m%n; } printf("最大公约数：%d\n",n);}
- = 0) { if (m%2 == 0) printf("%d is a positive even\n", m);Else printf("%d is a positive odd\n", m); }Else { if (m % 2 == 0) printf("%d is a negative even\n", m); Else printf("%d is a negative odd\n", m); }}">7.#include <stdio.h>Main(){ int m;Scanf("%d", &m); If (m >= 0) { if (m%2 == 0) printf("%d is a positive even\n", m);Else printf("%d is a positive odd\n", m); }Else { if (m % 2 == 0) printf("%d is a negative even\n", m); Else printf("%d is a negative odd\n", m); }}
- 8.#include <stdio.h>Main(){ int m; scanf("%d", &m); if (m >= 0) { if (m%2 == 0) printf("%d is a positive even\n", m); else printf("%d is a positive odd\n", m); } else { if (m % 2 == 0) printf("%d is a negative even\n", m); else printf("%d is a negative odd\n", m); }}
- = 0) { if (m%2 == 0) printf("%d is a positive even\n", m); else printf("%d is a positive odd\n", m); } else { if (m % 2 == 0) printf("%d is a negative even\n", m); else printf("%d is a negative odd\n", m); } 选项： A:A. -9 is a positive even B:B. -9 is a positive odd C:C. -9 is a negative even D:D. -9 is a negative odd">9.若运行时输入-9，则以下程序的运行结果是 int m; scanf("%d", &m); if (m >= 0) { if (m%2 == 0) printf("%d is a positive even\n", m); else printf("%d is a positive odd\n", m); } else { if (m % 2 == 0) printf("%d is a negative even\n", m); else printf("%d is a negative odd\n", m); } 选项： A:A. -9 is a positive even B:B. -9 is a positive odd C:C. -9 is a negative even D:D. -9 is a negative odd
-
10.求两个数m和n的最大公约数正确算法的伪代码是？选项： A:input m,nif m
- 11.阅读下面程序段,则执行后的结果为 选项： #include "stdio.h" main() { int m=4,n=2,k; k=fun(m,n); printf("%d ",k);} fun(int m,int n) { return(m*m*m-n*n*n);} A．64 B．8 C．56 D．0
-
12.【填空题】下列程序的输出结果为 #include
int fun3 (int m); int main (void) { int n; for (n=1; n< 10; n++) if (fun3(n) == 1) printf ( “ %d ” ,n); return 0; } int fun3(int m) { int i; if (m == 2|| m == 3) return 1; if (m < 2 || m%2 == 0) return 0; for(i=3;i -
13.编程计算并输出m×n阶矩阵的转置矩阵。其中，m和n的值由用户从键盘输入。已知m和n的值都不超过10。按要求在空白处填写适当的表达式或语句，使程序完整并符合题目要求。#include
#define M 10#define N 10void Transpose(int a[][N], int at[][M], int m, int n);void InputMatrix(int a[][N], int m, int n);void PrintMatrix(int at[][M], int n, int m);int main(){ int s[M][N], st[N][M], m, n; printf("Input m, n:"); scanf("%d,%d", &m, &n); InputMatrix(s, m, n); Transpose(s, st, m, n); printf("The transposed matrix is:\n"); PrintMatrix(st, n, m); return 0;}/* 函数功能：计算m*n矩阵a的转置矩阵at */void Transpose(int a[][N], int at[][M], int m, int n) { int i, j; for (i=0; i -
14.请给出程序的运行结果。#include
int Fun(int m){ static int n = 1; m /= 2; m = m * 2; if (m) { n *= m; return(Fun(m - 2)); } else return n;} int main(){ int a, i; for (i = 0; i < 2; i++) { a = Fun(4 + i); printf("%d\n", a); } return 0;}选项： A:00; B:6464; C:44; D:864 -
15.编程计算并输出m×n阶矩阵的转置矩阵。其中，m和n的值由用户从键盘输入。已知m和n的值都不超过10。按要求在空白处填写适当的表达式或语句，使程序完整并符合题目要求。#include
#define M 10 #define N 10 void Transpose(int a[][N], int at[][M], int m, int n); void InputMatrix(int a[][N], int m, int n); void PrintMatrix(int at[][M], int n, int m); int main() { int s[M][N], st[N][M], m, n; printf("Input m, n:"); scanf("%d,%d", &m, &n); InputMatrix(s, m, n); Transpose(s, st, m, n); printf("The transposed matrix is:"); PrintMatrix(st, n, m); return 0; } /* 函数功能：计算m*n矩阵a的转置矩阵at */ void Transpose(int a[][N], int at[][M], int m, int n) { int i, j; for (i=0; i
