C语言程序设计（第四版）谭浩强版 课后答案 第五章

2、

#include<stdio.h>
#include<math.h>
int main(){
        int sign = 1,count = 0;
        double pi = 0.0,n = 1.0,term = 1.0;
        while(fabs(term)>=pow(10,-6)){
                pi = pi +term;
                n = n+2;
                sign = -sign;
                term = sign/n;
                count++;
        }
        pi = pi*4;
        printf("pi = %10.8f\n",pi);
        printf("The count is:%d\n",count);
        return 0;
}

3、辗转相除法

#include<stdio.h>
int main(){
        int a,b,temp = 0;
        int x,y;
        scanf("%d%d",&a,&b);
        x = a;
        y = b;
        while(b){
                temp = a%b;
                a = b;
                b = temp;
        }
        printf("maxdiv:%d\n",a);
        printf("minmul:%d\n",x*y/a);
        return 0;
}

4、容易出现问题，判断语句中=一定是==

#include<stdio.h>
int main(){
        int a = 0;
        int b = 0;
        int c = 0;
        int o = 0;
        char ch;
        scanf("%c",&ch);
        while(ch!='\n'){
                if(ch>='A'&&ch<='z')
                        a++;
                else if((int)ch==32)
                        b++;
                else if (ch>='0'&&ch<='9')
                        c++;
                else o++;
                scanf("%c",&ch);
        }
        printf("The number of letter is:%d\n",a);
        printf("The number of blank is:%d\n",b);
        printf("The number of figure is:%d\n",c);
        printf("The number of other is:%d\n",o);
        return 0;
}

5、

#include<stdio.h>
#include<math.h>
int main(){
        int n,a=0,sum=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++){
                a=a+pow(10,i)*2;
                sum+=a;
        }
        printf("sum:%d\n",sum);
        return 0;
}
   

6、

#include<stdio.h>
int main(){
        int i,j=1,sum=0;
        for(i=1;i<=20;i++){
                j*=i;
                sum+=j;
        }
        printf("sum is:%d\n",sum);

        return 0;
}

7、

#include<stdio.h>
#include<math.h>
int main(){
        int sum1=0,sum2=0;
        float sum3=0;
        for(int i=1;i<=100;i++)
                sum1+=i;
        for(int i=1;i<=50;i++)
                sum2+=pow(i,2);
        for(int i=1;i<=10;i++)
                sum3+=1.0*1/i;
        printf("sum is :%0.2f\n",sum1+sum2+sum3);

        return 0;
}

8、水仙花数

#include<stdio.h>
int main(){
        int a,b,c;
        for(int i=100;i<1000;i++){
                a = i%10;
                b = i/10%10;
                c = i/100;
                if(a*a*a+b*b*b+c*c*c==i)
                        printf("it is a narcinum:%d\n",i);

        }

        return 0;
}

9、问题，第一遍sum放在int定义后面，所以结果为空，在计算每个数是不是完数时，都应该把sum归零

#include<stdio.h>
int main(){
        int i,j,sum;
        for(i=2;i<=1000;i++){
                sum = 0;
                for(j=1;j<i;j++){
                        if(i%j==0)
                                sum+=j;
                }
                if(sum==i){
                        printf("%d its factors are:",i);
                        for(j=1;j<i;j++){
                                if(i%j==0)
                                        printf("%d,",j);
                        }
                        printf("\n");
                }
        }
        return 0;
}

10、

#include<stdio.h>
int main(){
        float sum;
        int a = 2,b = 1,c;
        for(int i=1;i<=20;i++){
                sum+=(float)a/b;
                c=a;
                a=a+b;
                b=c;

        }
        printf("the sum:%f\n",sum);
        return 0;
}
        

11、

#include<stdio.h>
#include<math.h>
int main(){
        float sum=100,high;
        for(int i=1;i<10;i++){
                sum=sum+(1.0*100/pow(2,i))*2;
        }
        high = 0.1*100/pow(2,10);
        printf("sum:%f,high:%f\n",sum,high);
        return 0;
}
                                                                   

12、正好返回来求，确定循环几次。

#include<stdio.h>
int main(){
        int sum = 1,i;
        for(i=1;i<=9;i++){
                sum=(sum+1)*2;
        }
        printf("sum is :%d\n",sum);

        return 0;
}

13、 

#include<stdio.h>
#include<math.h>
int main(){
        float a,m,n;
        printf("please input:");
        scanf("%f",&a);
        m = a/2;
        while(1){
                n = m;
                m = 0.5*(n+a/n);
                if(fabs(n-m)<pow(10,-5))
                        break;
        }
        printf("the result is:%f\n",m);
        return 0;

}

14、熟悉牛顿迭代法定义

#include<stdio.h>
#include<math.h>
int main(){
        float x,x1=1.5,f,f1;
        for(x=0;fabs(x1-x)>=pow(10,-5);){
                x=x1;
                f=2*x*x*x-4*x*x+3*x-6;
                f1=6*x*x-8*x+3;
                x1=x-f/f1;
                printf("x1=%f\n",x1);

        }
        return 0;
}

15、

#include<stdio.h>
#include<math.h>
int main(){
        float a,a1,b,x=-10,y=10;
        a = 10;
        while(1){
                a1 = a;
                a = (x+y)/2;
                b = 2*a*a*a-4*a*a+3*a-6;
                if(b>0)
                        y=a;
                else
                        x=a;
                if(fabs(a1-a)<=pow(10,-5))
                        break;
        }
        printf("the result is:%f\n",a);
        return 0;
}

16、