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阿新 • • 發佈:2020-11-28
實驗1
一元二次方程的根,不能設計成以函式返回值的方式返回給主調函式
實驗2
#include <stdio.h> long long fac(int n); int main() { int i,n; printf("Enter n: "); scanf("%d", &n); for(i=1; i<=n; ++i) printf("%d! = %lld\n", i, fac(i)); return 0; } long long fac(int n) { static long long p = 1; printf("p = %lld\n", p); p = p*n; return p; }
#include<stdio.h> int func(int, int); int main() { int k=4,m=1,p1,p2; p1 = func(k,m) ; p2 = func(k,m) ; printf("%d,%d\n",p1,p2) ; return 0; } int func(int a,int b) { static int m=0,i=2; i+= m+1; m = i+a+b; return (m); }
區域性static變數只賦初值一次,以後每次呼叫函式時不再重新賦初值而保留上次函式呼叫結束時的值。
實驗3
#include <stdio.h> #define N 1000 int fun(int n,int m,int bb[N]) { int i,j,k=0,flag; for(j=n;j<=m;j++) { flag=1; for(i=2;i<j;i++) if(j%i==0) { flag=0; break; } if(flag!=0) bb[k++]=j; } return k; } int main(){ int n=0,m=0,i,k,bb[N]; scanf("%d",&n); scanf("%d",&m); for(i=0;i<m-n;i++) bb[i]=0; k=fun(n,m,bb); for(i=0;i<k;i++) printf("%4d",bb[i]); return 0; }
實驗4
#include <stdio.h> long long fun(int n); int main() { int n; long long f; while(scanf("%d", &n) != EOF) { f = fun(n); printf("n = %d, f = %lld\n", n, f); } return 0; } long long fun(int n){ long long k; if(n==0) k=0; if(n==1) k=1; if(n>=2) k=2*fun(n-1)+1; return k; }
實驗5
#include <stdio.h> void draw(int n, char symbol); #include <stdio.h> int main() { int n, symbol; while(scanf("%d %c", &n, &symbol) != EOF) { draw(n, symbol); printf("\n"); } return 0; } void draw(int n, char symbol){ int x,y,z; for(x=1;x<=n;x++){ for(y=1;y<=n-x;y++){ printf(" "); } for(z=1;z<=2*x-1;z++){ printf("%c",symbol); }printf("\n"); } }