main() {int n=0,m=1,x=2;if(!n)x=-1;if(m)x=x-2;if(x)x=-3;printf("%d\n",x);}求运行后的结果,刚开始学,给说详细一点,为什么得出这个数,我知道是-3

#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 #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 #include int main() { int a[150];int m,n,i,j,t; while(scanf(%d %d,&m ,&n)!=EOF) {printf(#includeint main(){int a[150];int m,n,i,j,t;while(scanf(%d %d,&m ,&n)!=EOF){printf(%d~%d prime include:,m,n);t=0;for ( i=m+1;i #includeint a(int x);int b(int y);int main(){int m,n;n=6;m=a(n);printf(%d ,m);return 0;}int a(int x){x=b(x+3);return x;}int b(int y){return y+1;} 为什么输入1得到了2 百思不得其解!//#include int math(n){if (n>=3)return math(n-1)+math(n-2);else if(n=2)return 2;else if(n=1)return 1;}int sum (int n){int m=0;for (;n>0;n--){m+=math(n);}return m;}main(){int i=0;printf(30个数和为：% #include void main() {int a=1,m,n,j,s=0;#includevoid main(){int a=1,m,n,j,s=0;scanf(%d,&n);for(j=1;j #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/n整除 程序有错 编写程序 计算组合数的公式#include int main(void){double fact(int n,int m);int i,p,m,n;double result,product,case;printf(Enter n,m:);scanf(%d%d,&n,&m);for(i = 0;i syntax error :identifier 'n'#includeint i,n,s;int proone(int i,n,s=0);int protwo(int i,n,s=0);void main( ){double s1,s2;int m,n;cout 计算1-2+3-4+……+/-n的C++程序int main(){int sum(int);int n;coutn;if (n%2==0)cout 用C语言实现一个函数,函数功能是返回一个数能被另一个数整除的次数#include#includeint count(int n,int m){int i;int num=0;for(i=n;i>1;i/m){if(i%m==0){num++;}}return num;}int main(){int a;scanf(%d,&a);int t=count(a,2);printf 编写一个函数,求一组数中的最大值和最小值源程序：int min;int maximum(n)；main(){ int i,n[10],m;clrscr();for(i=0；i main() { int n,s; n=1; s=0; for(;n C语言 运算最小公倍数问题#include int fun(int x,int y,int z){ int j,t,n,m;// while (t!=0 || m!=0 || n!=0) { j=j+1; t=j%x; m=j%y; n=j%z; } return j;}main(){ int x1,x2,x3,j; printf(Input x1,x2,x3:); scanf(%d%d%d,&x1,&x #include int ZDGYS(int m,int n) { int r; scanf(%d%d,&m,&n); r=m%n; while(r!=0) { m=n; n=#includeint ZDGYS(int m,int n){int r;scanf(%d%d,&m,&n);r=m%n;while(r!=0){m=n;n=r;r=m%n;}return n;}main(){printf(%d,r);}求m,n的最大公约数 看看哪有 #include int main( ) { int a ,b,m ,n ; m = n = 1 ; scanf( %d%d ,&a ,&b) ;#includeint main( ){int a ,b,m ,n ;m = n = 1 ;scanf( %d%d ,&a ,&b) ;do{if(a > 0 ){m= 2 * n ;b++;}else{n= m + n ;a+= 2 ;b++;}}while( a ==b ) ;printf( m=%d n=%d ,m ,n ) ;ret #include int main() { int a[100]; int n,i,j,k,t,m; scanf(%d %d ,&m,&n);#includeint main(){int a[100];int n,i,j,k,t,m;scanf(%d %d ,&m,&n);scanf(%d,&a[i]);for(i=0;i #include void f(int *p,int*q); main() {int m=1,n=2,*r=&m; f(r,&n);printf(%d,%d,m,n);}#include void f(int *p,int*q);main(){int m=1,n=2,*r=&m;f(r,&n);printf(%d,%d,m,n);}void f(int *p,int *q){p=p+1;*q=*q+1;}运行后的结果是A 1,3 B 1,2 说说为