有以下程序 #include <stdio.h> #define N 5 #define M N+1 #define f(x) (x*M) main() { int i1,i2; i1=f(2); i2=f(1+1); printf("%d%d/n",i1,i2); } 程序的运行结果是
A.12 12
B.11 7
C.11 11
D.12 7
[单选题]有以下程序includedefineN5defineMN£«1definef(x)(x*M)main(){inti1,i2;i1=f(2);i2=有以下程序 #include <stdio.h> #define N 5 #define M N+1 #define f(x) (x*M) main() {int i1,i2; i1=f(2); i2=f(1+1); printf("%d%d/n",i1,i2); } 程序的运行结果是A.12 12B.11 7C.11 11D.12 7
[单选题]有以下程序#include <stdio.h>#define N 5#define M N+1#define f(x)(x*M)main(){ int i1,i2;i1=f(2);i2=f(1+1);printf("%d %d/n",i1,i2);}程序的运行结果是A.12 12B.11 7C.11 11D.12 7
[单选题]有如下程序:#define N 2#define M N+1#define NUM 2*M+1main(){int i;for(i=1;i<=NUM;i++)printf("%d\n",i);}该程序中的for循环执行的次数是A.5B.6C.7D.8
[单选题]有如下程序: #define n 2 #define m N+1 #define NUM 2*m+1 main() { int i; for(i=1;i<=NUM;i++)printf("%d/n",i); } 该程序中的for循环执行的次数是______。A.5B.6C.7D.8
[单选题]有如下程序: #define n 2 #define m N+1 #define NUM 2*m+1 main() { int i; for(i=1;i<=NUM;i++)printf("%d/n",i); } 该程序中的for循环执行的次数是_____。A.5B.6C.7D.8
[单选题]有以下程序#include<stdio.h>#define N 4void fun(int a[][N],int b[]){ int i;for(i=0;i<N;i++)b[i]=a[i][i]-a[i][N-1-i];}main(){ int x[N][N]:{{1,2,3,4},{5,6,7,8},{9,10,11,12},{13,14,15,16}},y[N],i;fun(x,y);for(i=0;i<N;i++) prinff("%d,",y[i]);prin
[单选题]有以下程序:includeusing namespace std;definePl 3.14Class Point{private:int x,y有以下程序: #include<iostream> using namespace std; #definePl 3.14 Class Point {private: int x,y; public: Point(int a,intB) {X=a; y:b;} int getx() <return x;} int gety() {return y;}}
[单选题]有以下程序:include include main(){char *p[10]={"abc","aabdfg","dcdbe"有以下程序: #include <stdio.h> #include <string.h> main() { char *p[10]={"abc","aabdfg","dcdbe","abbd","cd"}; printf("%d/n",strlen(p[4])); } 执行后的输出结果是( )。A.2B.3C.4D.5
[单选题]有以下程序:includedefine N 5define M N£«1define f(x)(x*M)main(){int i1,i2;i1=f有以下程序: #include<stdio.h> #define N 5 #define M N+1 #define f(x)(x*M) main() {int i1,i2; i1=f(2); i2=f(1+1); printf("%d%d",i1,i2); } 程序的运行结果是( )。A.12 12B.11 7C.11 11D.12 7
[单选题]阅读下列程序段,则程序的输出结果为 #include"stdio.h" #defineM(X,Y)(X)*(Y) #defineN(X,Y)(X)/(Y) main() {f int a=5,b=6,c=8,k; k=N(M(a,b),c); printf("%d/n",k);}A.3B.5C.6D.8