第4章 习题解答

1、4.1 编写一个程序，输入A，B，C三个学生的考试分数，输出分数居中的那个学生名（A、B或C）及考试分数。#include <stdio.h>int main() float a, b, c; printf("Please enter the score of A:n"); scanf("%f", &a); printf("Please enter the score of B:n"); scanf("%f", &b); printf("Please enter the sco

2、re of C:n"); scanf("%f", &c); if (a-b)*(a-c) <= 0) printf("A gets the score %.1f", a); else if (b-a)*(b-c) <= 0) printf("B gets the score %.1f", b); else printf("C gets the score %.1f", c); return 0;4.2 编程求一元二次方程ax2+bx+c=0的根，系数a,b,c从终端输入。应考虑实根、复

3、数根和无解的情况。#include<stdio.h>#include<math.h>int main(void)float a, b, c, d, x1, x2;printf("Please key in 3 numbers:");scanf("%f%f%f", &a, &b, &c);if(a != 0)d = b*b - 4*a*c;if(d > 0)x1 = (sqrt(d) - b) / (2*a);x2 = -(sqrt(d) + b) / (2*a);printf("Two re

4、al results:n%.6fn%.6fn", x1, x2);else if(d = 0)x1 = -b/(2*a);printf("One real result:n%.6fn", x1);elsex1 = -b/(2*a);x2 = sqrt(-d)/(2*a);printf("Two virsual results:n%.3f+%.3fin%.3f-%.3fin", x1, x2, x1, x2);else if(b != 0)x1 = -c/b;printf("One real result:n%.6fn", x

5、1);else if(c!=0)printf("Error!No result.n");elseprintf("Anyone is result.n");return 0;4.3 输入一个日期（年、月、日），计算并输出该日期是这一年的第几天。switch方法：int main() int y, m, d, days; printf("Enter year:"); scanf("%d", &y); printf("Enter month:"); scanf("%d",

6、&m); printf("Enter day:"); scanf("%d", &d); if(m<1 | m>12) printf("Input error!n"); else days = d; switch(m) case 12: days += 30; case 11: days += 31; case 10: days += 30; case 9: days += 31; case 8: days += 31; case 7: days += 30; case 6: days += 31; case

7、 5: days += 30; case 4: days += 31; case 3: days += (y%4=0 && y%100!=0 | y%400=0 ? 29 : 28); case 2: days += 31; printf("%dn", days); return 0;函数方法：#include <stdio.h>int mdays(int y, int m)if (m=2) return y%4=0 && y%100!=0 | y%400=0 ? 29 : 28;else if (m=4 | m=6 | m=9

8、| m=11) return 30;else return 31;int main()int y, m, d, days; printf("Enter year:"); scanf("%d", &y); printf("Enter month:"); scanf("%d", &m); printf("Enter day:"); scanf("%d", &d); if(m<1 | m>12 | d>mdays(y, m) printf(

9、"Input error!n"); else days = d; while(m>1) days += mdays(y, m-1); m-; printf("%dn", days); return 0;4.4 假设工资税金按以下方法计算：x < 1000元，不收取税金；1000x <2000，收取5%的税金；2000 x < 3000，收取10%的税金；3000 x < 4000，收取15%的税金；4000x<5000，收取20%的税金；x>5000，收取25%的税金。输入工资金额，输出应收取税金额度，要求分别用

10、if语句和switch语句来实现。if方法：#include <stdio.h>int main()float x = 0; printf("input the salary:"); scanf("%f",&x); if(x<0) printf("Input error!n"); else if(x<1000) printf("0.00n"); else if(x<2000) printf("%.2fn",x*0.05); else if(x<3000

11、) printf("%.2fn",x*0.1); else if(x<4000) printf("%.2fn",x*0.15); else if(x<5000) printf("%.2fn",x*0.2); else printf("%.2fn",x*0.25); return 0;switch方法：#include <stdio.h>int main() float x ; int xCase = 0; printf("input the salary:"); scan

12、f("%f",&x); if(x<0) printf("Input error!n"); else xCase = (int)(x/1000.0); switch(xCase) case 0: printf("0.00n"); break; case 1: printf("%.2fn",x*0.05); break; case 2: printf("%.2fn",x*0.1); break; case 3: printf("%.2fn",x*0.15); bre

13、ak; case 4: printf("%.2fn",x*0.2); break; default: printf("%.2fn",x*0.25); return 0;4.5 输入两个实数和一个四则运算符（+、-、*、/），根据运算符执行相应的运算并输出运算结果，分别用if语句和switch语句来实现。if方法：#include<stdio.h>int main(void)float a, b, result;char c;printf("Please key in 2 numbers:");scanf("%f%

14、f", &a, &b);getchar();printf("Please key in 1 operator:");scanf("%c", &c);if(c = '+')result = a+b;printf("%f+%f=%.6fn", a, b, result);else if(c = '-')result = a-b;printf("%f-%f=%.6fn", a, b, result);else if(c = '*')resu

15、lt = a*b;printf("%f*%f=%.6fn", a, b, result);else if(c = '/')if(b!=0)result = a/b;printf("%f/%f=%.6fn", a, b, result);elseprintf("Error!No result.n");elseprintf("Error!No result.n");return 0;switch方法：#include<stdio.h>int main(void)float a, b, res

16、ult;char c;printf("Please key in 2 numbers:");scanf("%f%f", &a, &b);getchar();printf("Please key in 1 operator:");scanf("%c", &c);switch(c)case '+':result = a+b;printf("%f+%f=%.6fn", a, b, result);break;case '-':result = a

17、-b;printf("%f-%f=%.6fn", a, b, result);break;case '*':result = a*b;printf("%f*%f=%.6fn", a, b, result);break;case '/':if(b!=0)result = a/b;printf("%f/%f=%.6fn", a, b, result);elseprintf("Error!No result.n");break;default:printf("Error!No r

18、esult.n");return 0;4.6 统计输入正文中空格字符、制表符和换行符的个数。#include<stdio.h>int main(void)int a = 0, b = 0, c = 0;int ch;while(ch=getchar() != EOF)if(ch = ' ')a+;else if(ch = 't')b+;else if(ch = 'n')c+;printf("The number of space, tab, enter is :%d, %d, %dn", a, b, c)

19、;return 0;4.7 编写一个程序，将输入的一行字符复制到输出，复制过程中将一个以上的空格字符用一个空格代替。#include <stdio.h>int main() char c; int flag = 0; do c = getchar(); if(c!=' ') putchar(c); flag = 0; else if(!flag) putchar(c); flag = 1; while(c != 'n'); return 0;4.8 将输入的正文复制到输出，复制过程中删去每个输入行的前导空格。#include<stdio.h&g

20、t;int main(void)int ch, flag, i=0;char s800;while(ch=getchar() != EOF && i < 799)si+ = (char)ch;si = 0;flag = 1;printf("nn");for(i=0; si!=0; i+)if(!(flag && si = ' ')putchar(si);if(si = 'n')flag = 1;elseflag = 0;return 0;4.9 输出斐波拉契数列的前n项。例如：11235813n由终端输入

21、，n>=20，每行输出8个数。计算公式为：a1=1,a2=1,an=an-1+an-2(n>3)。#include<stdio.h>int main(void)int n, i, fib100;printf("Please key in a number:");scanf("%d", &n);fib0 = 1;printf("%6d", fib0);fib1 = 1;printf("%6d", fib1);for(i=2; i<n; i+)fibi = fibi-1 + fib

22、i-2;printf("%6d", fibi);if(i+1)%8 = 0)putchar('n');return 0;4.11 编写一个程序，输入两个整数，求它们的最大公约数和最小公倍数。#include<stdio.h>int main() int a,b,num1,num2,temp; printf("Input a & b:"); scanf("%d%d",&num1,&num2); a=num1; b=num2; if(a>b) temp=a; a=b; b=temp

23、; while(b!=0) temp=a%b; a=b; b=temp; printf("The GCD of %d and %d is: %dn",num1,num2,a); printf("The LCM of them is: %dn",num1*num2/a); return 0;4.12 输入两个整数，判断它们是否为互质数并输出判断结果。/*公因数只有1的两个非零自然数,叫做互质数。举例：2和3，公因数只有1，为互质数*/#include<stdio.h>int main(void)int m, n, i, min;printf(&

24、quot;Please key in two number:");scanf("%d%d", &m, &n);min = m>n ? n : m;for(i=2; i<=min; i+)if(!(m%i)if(!(n%i)break;if(i>min)printf("Yesn");elseprintf("Falsen");return 0;4.13 编写一个程序，验证哥德巴赫猜想：任何充分大（>=4）的偶数都可以用两个素数之和表示，将4100中所有偶数分别用两个素数之和的形式输出。（例

25、如：4=2+2,100=3+97）#include <stdio.h>#include <math.h>int Primes(int x); /判断素数函数int main() int i;int num; for(num=4;num<=100;num+=2) for(i=1;i<=num/2;i+) if(Primes(i) && Primes(num-i) printf("%d=%d+%dn", num, i, num-i); return 0; int Primes(int x) int i, j; j = sqrt(x); for( i=2; i<=j; i+) if(x%i=0) break; if( i>j ) return 1; else return 0;4.15 输出10000以内所有这样的完全平方数：a2 = b2*10 + c2。例如，361=192=62*10+12。#include<stdio.h>#define N 10000int main(void)int a, b, c, l, m, n;for(a=1; (l