实验四
#include <math.h> #include <stdio.h> void solve(double a, double b, double c); int main() { double a, b, c; printf("Enter a, b, c: "); while(scanf("%lf%lf%lf", &a, &b, &c) != EOF) { solve(a, b, c); printf("Enter a, b, c: "); } return 0; } void solve(double a, double b, double c) { double x1, x2; double delta, real, imag; if(a == 0) printf("not quadratic equation.\n"); else { delta = b*b - 4*a*c; if(delta >= 0) { x1 = (-b + sqrt(delta)) / (2*a); x2 = (-b - sqrt(delta)) / (2*a); printf("x1 = %f, x2 = %f\n", x1, x2); } else { real = -b/(2*a); imag = sqrt(-delta) / (2*a); printf("x1 = %f + %fi, x2 = %f - %fi\n", real, imag, real,imag); } } }
根不能一次性以函数返回值的形势返还给主函数,每个函数返回值只有一个,若要返还可以存入数组或者将x1,x2分为两个子函数分别计算和返还
#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; p = p*n; return p; }
插入前:
插入后:
#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); }
理论分析与实验结果一致
局部static变量的特性:
该变量在全局数据区分配内存;
静态局部变量一般在声明处初始化,如果没有显式初始化,会被程序自动初始化为0;
它始终驻留在全局数据区,直到程序运行结束。但其作用域为局部作用域,当定义它的函数或语句块结束时,其作用域随之结束;
#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; }
#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 = 1; else k = 2*(fun(n-1)+1); k = k-1; return k; }
#include <stdio.h> void fun(int n,int a[]); int main() { int n; while(scanf("%d", &n) != EOF) { int a[1000]={1,0}; fun(n,a); printf("\n"); } return 0; } void fun(int n,int a[]){ int i,j,k=0,c=0; for(i=1;i<=n;i++){ for(j=0;j<=k;j++){ a[j]=a[j]*2+c; c=a[j]/10; a[j]=a[j]%10; } if(c){ a[++k]=c; } c=0; } printf("n = %d,f =", n); for(i=k;i>=0;i--){ if (i==0) a[i]-=1; printf("%d",a[i]); } }
算法思路:因为数太大,所以不能直接输出计算;
设立一个足够大的数组,每一个元素储存一位数,判定低位数超过十则下一位加一;
最后将个位的数组减一,因为不存在个位数字为零需要借位,所以直接按位数依次输出
#include <stdio.h> void draw(int n, char symbol); 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,j,m,k; for(i=1;i<=n;i++){ j=n-i; m=2*i-1; for(k=1;k<=j;k++) printf(" "); for(k=1;k<=m;k++){ printf("%c",symbol); if(k==m) printf("\n"); } } }