Chapter Four
3. 写出下面各逻辑表达式的值。设a=3,b=4,c=5。
（1）a + b > c && b == c
（2）a || b + c && b - c
（3）!(a > b) && !c || 1
（4）!(x = a) && (y = b) && 0
（5）!(a + b) + c - 1 && b + c / 2
(1) 0
3+4>5 gives priority to 3+4 to get the result 7, so the result of 7>5 is true; 4==5 is false, one true and one false logic and the final result is false.
(2) 1
priority arithmetic operation 4+5 gets 7, if it is not 0, it is true, 4-5 gets -1, if it is not 0, it is true, then logic and judgment, and finally logic or judgment.
(3) 1
! The highest priority, !(3>4) the final result is true, !5 is false; followed by &&, true && false to get false, and finally ||, 1 is true, false or true is true (4) 0 && has the
lowest
priority It is the last logical operation, so no matter what, in the end && 0, it must be false
(5) 1
will first negate (a+b) in vs to get 0, 0+5-1 and the result is 4, so it is finally true ( Compilers under different platforms may give priority to arithmetic operations, and the result of negation will be false)
8. 给出一百分制成绩,要求输出成绩等级’A’、‘B’、‘C’、‘D’、‘E’。
90分以上为’A’,8089分为’B’,7079分为’C’ ,60~69分为’D’ ,60分以下为’E’。
switch - case
#include<stdio.h>
#include<stdlib.h>
int main()
{
float score;
char grade;
printf("请输入学生成绩：");
scanf("%f", &score);
while(score > 100 || score < 0)
{
printf("\n输入错误,请重新输入");
scanf("%f", &score);
}
switch((int)(score/10))
{
case 10:
case 9: grade = 'A';break;
case 8: grade = 'B';break;
case 7: grade = 'C';break;
case 6: grade = 'D';break;
case 5:
case 4:
case 3:
case 2:
case 1: grade = 'D';
}
printf("成绩是%5.1f,相应的等级为%c\n",score,grade);
return 0;
}
if - else
if (score >= 90) {
printf("A\n");
}else if (score >= 80 && score < 90) {
printf("B\n");
}else if (score >= 70 && score < 80) {
printf("C\n");
}else if (score >= 60 && score < 70) {
printf("D\n");
}else {
printf("E\n");
}
chapter Five
#include <stdio.h>
#include <math.h>
int main()
{
//n为a的个数
int n;
double a, prev_sum = 0.0, total_sum = 0.0;
printf("请输入a的值以及n的值: ");
scanf("%lf %d", &a, &n);
//循环n次求总和
for (int i = 0; i < n; i++)
{
prev_sum += a * pow(10, i);
total_sum += prev_sum;
}
printf("总和为：%lf\n", total_sum);
return 0;
}
#include<stdio.h>
int main()
{
double total_sum = 0,t = 1;
for(int i = 1; i <= 20; i++)
{
t = t * i;
total_sum = total_sum + t;
}
printf("1~20每个数字阶乘总和为：%lf\n",total_sum);
return 0;
}
#include <stdio.h>
int main()
{
//a表示百位数字，b表示十位数字，c表示各位数字
int a, b, c;
for (int i = 100; i < 1000; i++)
{
a = i / 100;
b = (i / 10) % 10;
c = i % 10;
if (a * a * a + b * b * b + c * c * c == i)
{
printf("%d\n", i);
}
}
return 0;
}
#include<stdio.h>
int main()
{
int data, fator, sum; /* data表示要判断的数，fator表示因子，sum表示因子之和*/
for (data = 2; data <= 1000; data++)
{
//1是所有整数的因子，所以因子之和从1开始
sum = 1;
for (fator = 2; fator <= data / 2; fator++)
{
/* 判断data能否被fator整除，能的话fator即为因子 因子不包括自身 */
if (data % fator == 0)
{
sum += fator;
}
}
// 判断此数是否等于因子之和 */
if (sum == data)
{
printf("%d its factors are 1, ", data);
for (fator = 2; fator <= data / 2; fator++)
{
if (data % fator == 0)
{
printf("%d, ", fator);
}
}
printf("\n");
}
}
return 0;
}
#include <stdio.h>
int main()
{
int i, n = 20;
double a = 2, b = 1, s = 0, t;
for (i = 1; i <= n; i++)
{
s = s + a / b;
//记录前一项分子
t = a;
//前一项分子与分母之和为后一项分子
a = a + b;
//前一项分子为后一项分母
b = t;
}
printf("sum=%16.10f\n", s);
return 0;
}
#include <stdio.h>
#include <math.h>
int main()
{
float a, x0, x1;
printf("请输入一个正数: ");
scanf("%f", &a);
x0 = a / 2;
x1 = (x0 + a / x0) / 2;
do
{
x0 = x1;
x1 = (x0 + a / x0) / 2;
} while (fabs(x0 - x1) >= 1e-5);
printf("[%f] 的平方根为 [%f]\n", a, x1);
return 0;
}
