main() { int n=4; while(n--) printf("%d",--n); }

main() { int n=4; while(n--) printf(%d,--n); } 计算1-2+3-4+……+/-n的C++程序int main(){int sum(int);int n;coutn;if (n%2==0)cout 请问下面这个程序怎么解释 main() {int n=4; while(n--)printf(%d,--n); } void Function() { printf(Call Function! ); } int main() { void (*p)(); *(int*)&p=(int)Function; #include int call(int s[],int y) { static int t=3; y=s[t--]; return(y); } void main() { in#includeint call(int s[],int y){static int t=3;y=s[t--];return(y);}void main(){int a[]={3,8,4,2},n,x=0;for(n=0;n #define N 20 fun(int a[],int n,int m) {int i; for(i=m;i>n;i--)a[i+1]=a[i]; return m; } void main() #define N 20fun(int a[],int n,int m){int i;for(i=m;i>n;i--)a[i+1]=a[i];return m;}void main(){ int i,a[N]={1,2,3,4,5,6,7,8,9,10};fun(a,0,N/2);for(i=0;i main() { int n; clrscr(); n=n+n; printf(n=%d,n) } int fm(int a,int b) {if(b==1)return a; else return a+fm(a,b-1); } main( ) { printf(%d ,fm(4,3)) int fm(int a,int b) {if(b==1)return a; else return a+fm(a,b-1); } main( ) { printf(%d ,fm(4,3) main( ) 　{ int n; ( n=6*4,n+6),n*2; printf(“n=%d ”,n); } 　此程序的输出结果是? 要解析 ACM斐波那契数列超时我一共编写了两个程序,但是在学校OJ(南邮)上还是超时了,程序本身没有错,但是占用内存太大,希望个位帮我改进,如下:1#include stdio.hint main(){int s=0,f=1,F=0,n;do{scanf(%d,&n);}wh void main() {int x=4;if(x++>=5)printf(%d/n,x);else printf(%d/n,x--); #include void InSertSort(int R[],int n);特别这一行void main(){int arr[]={0,1,2,3,4,5,6,7,8,9};int n=9;InSertSort(arr,10);}void InSertSort(int R[],int n)特别这一行{int i,j;int temp;for(i=1;i=0&&temp C++计算阶乘n#include #include #include #include using namespace std;int main(){int sum = 1,d;int n;cin >> n;for(int i=1;i #include int gcd(int m,int n) { if(m%n==0) printf(%d ,n); else gcd(n,m%n); } main() { i#includeint gcd(int m,int n){if(m%n==0) printf(%d ,n);elsegcd(n,m%n);} main(){int m,n;scanf(%d%d,&m,&n);printf(%d,gcd(n,m%n));}求m整除n 计算圆周率的C语言程序#include int main (){int n;double s=1.0;for (n=1;n #include main() {double x=1.0; int n; for(n=0;n main() { int n,s; n=1; s=0; for(;n