一個菜鳥的社工之旅
阿新 • • 發佈:2020-12-02
實驗一
不可以 一元二次方程有兩個根,而函式返回值只能返回一個值。
實驗二
#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) { staticlong 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) { staticint m=0,i=2; i += m+1; m = i+a+b; return (m); }
總結 static區域性變數只被初始化一次,下一次依據上一次結果值
實驗三
#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==1) 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; }
實驗四
#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){ if(n==0) return 0; if(n>0) return 2*fun(n-1)+1; }
實驗五
#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 i, a, b, c, d; a = n; for (i = 1; i <= n; i++) { d = 2 * i - 1; for (b = 1; b <= (a - i); b++){ printf(""); } for (c = 1; c <= d; c++) { printf("%c", symbol); } printf("\n"); } }