include <stdio.h>void main(){int a,b;scanf("%d%d",&a,&b);if(a%b==0) printf("%d能被%d整除。",a,b);else printf("%d不能被%d整除。",a,b);}运行示例：判断y能被x整除：y%x==0
} include <stdio.h>int duihuan(int* pa, int* pb){ int i = 0; i = *pa; *pa = *pb; *pb = i; return 0;}int main(){ int a = 0; int b = 0; scanf("%d %d", &a, &b); printf("a=%d,b=%d\n", a, b); duihuan(&a, &b)...
int main(){ int a,b,sum;//一、定义变量,int代表数据类型——整数,sum求和;printf("请输入两个整数：");scanf("%d %d",&a,&b);//二、输入数据,%d对输入类型的限制代表——整数，并把这俩个数分别给&a和&b两个变量，&取地址运算符;sum=a+b;//三、数据处理,把a、b的和赋值给sum;...
return 0;} 2.用数组求解：include<stdio.h> int main(){ int a,b,c,t;printf("请输入三个整数:\n");scanf("%d\t%d\t%d", &a, &b, &c);if (a > b){ t = a;a = b;b = t;} if (a > c){ t = a;a = c;c = t;} if (b > c){ t = b;b = c;c =...
include<stdio.h>void main(){int a,b,ave;printf("输入两个整数： ');scanf("%d %d",&a,&b);ave=(a+b)/2;printf("平均值为： %d",ave);}include<stdio.h>int main(){int a,b;float ave;printf("Please input tow numbers:");scanf("%d%d",&a,&b);ave=1.0*(a+b)/2;...
int main() //主函数 { int flag, a,b; //flag标志是否为质数 这里设定b>a printf("请依次输入a,b两个整数，注意a<b,a和b用空格分开\n");scanf("%d %d",&a,&b);for (int i=a;i<=b;i++) //从a开始到b之间的所有数 { flag=0; //初始化，为质数 for (int j...
include<stdio.h> int main(){ int a,b;float x,y;char c1,c2;scanf("%d%d",&a,&b);//不要添加多余字符 scanf("%f%f",&x,&y);//float类型用%f getchar(); // 添加此句接收上一句按下的回车符 scanf("%c%c",&c1,&c2);//输入这两字符时需连在一起 输入 printf("%d ...
include<stdio.h>void main() { int a,b,i,j,k; scanf("%d%d",&a,&b); for ( i=a;i<=b;i++ ) { for ( j=2,k=1;j<=i/2;j++ ) if ( i%j==0 ) { k=0; break; } if ( k!=0 ) printf("%d ",i); } printf("\n");} ...
x:y;}void main(){int a,b,c,mini;printf("please input first num:\n");scanf("%d",&a);printf("please input second num:\n");scanf("%d",&b);printf("please input third num:\n");scanf("%d",&c);mini=min(a,min(b,c));printf("the minimum is %d\n",mini);}...
如果是a:b; c:t1;则程序是求得最小值；如果是b:a; t1:c;则程序是求得最大值；如果题目没有要求，b:a t1:c 也是正确的；三目运算符a<b?a:b;的意思是，如果a<b成立，则该式结果为a；否则结果为b；满足