第一次作业
第一题
例4.8
#include<stdio.h>
int main()
{
int year, leap; //自定义和是否闰年
printf("输入年份：");
scanf("%d", &year); //给变量year赋值
if (year % 4 == 0) //判断年份是否被4整除，{}内为可整除后命令
{
if (year % 100 == 0) //判断是否被100整除，{}内为可整除后命令
{
if (year % 400 == 0) //年份被100整除，判断是否被400整除
leap = 1; //年份被400整除
else leap = 0; //年份不被400整除
}
else leap = 1; //年份不被100整除
}
else leap = 0; //年份不被4整除
if (leap) printf("%d是闰年.\n", year);
else printf("%d不是闰年.\n",year);
例5.1
#include<stdio.h>
int main()
{
int i = 1, sum = 0;
while (i <= 100)
{
sum = i + sum;
i++;
}
printf("1+2+...+100=%d", sum);
return 0;
}
例5.2
#include<stdio.h>
int main()
{
int i = 1, sum = 0;
do
{
sum = sum + i;
i++;
} while (i <= 100);
printf("1-100总和为%d.\n", sum);
}
例5.7
#include<stdio.h>
#include<math.h>
int main()
{
double pi = 0, k = 1, n = 1;
int z = 0;
do
{
z = z + 1;
pi = pi + k / n;
k = -k;
n = n + 2;
} while (fabs(k / n) >= pow(10, -6));
pi = pi * 4;
printf("圆周率为%10.8f\n.\n并且循环体一共循环了%d次", pi,z);
}
例5.9
#include<stdio.h>
int main()
{
int n,i;
printf("输入一个大于3的数字：");
scanf("%d",&n);
for (i = 2; i <= n; i++)
if (n % i == 0)break;
if (i < n)printf("%d不是素数", n);
else printf("%d是素数",n);
}
第二题
#include<stdio.h>
#include<math.h>
#include<conio.h> //第二行与第三行引用含有cos的函数
int main(void)
{
double m, n; //定义变量m与n，浮点计算，数据类型用“%lf”
printf("需要计算的角度(弧度制)="); //输入前要
scanf("%lf.\n", &m); //输入需要计算的角度
n = cos(m); //计算输入的角度的cos值
printf("cos(%lf)是 %lf.\n", m, n); //输出计算的角度的cos值
return 0;
}
第三题
#include<stdio.h>
#include<math.h>
int main(void)
{
float d = 300000, p = 6000, r = 0.01, m;
m = log10(p / (p - d * r)) / log10(1 + r);
printf("还清需要月数为%.1f",m);
return 0;
}
第二次作业
第一题
#include<stdio.h>
int main()
{
double a, b, ave;
printf("请输入两个数字且中间以空格隔开.\n");
scanf("%lf %lf",&a,&b);
ave = (a + b) / 2;
printf("%lf和%lf的平均值为%lf", ave);
}
第二题
#include<stdio.h>
int main()
{
printf("请依次输入abcdefg（支持小数）.\n");
double a, b, c, d, e, f, g, x;
printf("a="); scanf("%lf.\n", &a);
printf("b="); scanf("%lf.\n", &b);
printf("c="); scanf("%lf.\n", &c);
printf("d="); scanf("%lf.\n", &d);
printf("e="); scanf("%lf.\n", &e);
printf("f="); scanf("%lf.\n", &f);
printf("g="); scanf("%lf.\n", &g);
x = a + b * (c - d) / e * f - g;
printf("a + b * (c - d) / e * f - g=%lf", x);
return x;
}
第三题
#include <stdio.h>
int main()
{
double i,j,a,b,c,d,e,f,g;
scanf("%lf,%lf",&i,&j);
a=i+1;
b=j+1;
c=i++;
d=j++;
e=++i;
f=++j;
g=(i++)+(++j);
printf("i+1=%lf,j+1=%lf,i++=%lf,j++=%lf,++i=%lf,++j=%lf,(i++)+(++j)=%lf\n",a,b,c,d,e,f,g);
return 0;
}
第四题
#include<stdio.h>
#include<math.h>
int main()
{
printf("按提示输入对应数据，回车确认");
double year, money, rate, interest;
printf("存款金额为"); scanf("%lf.\n", &money);
printf("存款年数为"); scanf("%lf.\n", &year);
printf("年利率为"); scanf("%lf.\n", &rate);
interest = (money * pow(1 + rate, year) - money)/10;
printf("税前利息为%0.2f\n", interest);
return 0;
}
第五题
#include <stdio.h>
int main()
{
double c,f;
scanf("%lf",&f);
c=5*(f-32)/9;
printf("c=%lf\n",c);
return 0;
}
韩信点兵
#include<stdio.h>
#include<math.h>
int main()
{
int n = 0;
do {
n = n + 1;
}
while (n%5!=1||n%6!=5||n%7!=4||n%11!=10);
printf("韩信至少拥有的士兵数为%d",n);
return 0;
}
第三次作业
实验4
#include<stdio.h>
int main()
{
int a , b , c , d , k1, k2, k;
printf("求最大值，请输入四个整数");
scanf("%d%d%d%d", &a, &b, &c, &d);
int max(int x, int y);
k1 = max(a, b);
k2 = max(c, d);
k = max(k1, k2);
printf("输入的四个数中最大的是 % d", k);
return (k);
}
int max(int x, int y)
{
int m ;
if (x >= y) m = x;
else m = y;
return m;
}
第四次作业
例题3.4
#include<stdio.h>
#include<math.h>
int main()
{
printf("请输入满足三角形三遍关系的三条边的长度:");
double a, b, c, s, area;
scanf("%lf%lf%lf", &a, &b, &c);
s = (a + b + c) / 2;
area = sqrt(s * (s - a) * (s - b) * (s - c));
printf("三角形的面积为%f", area);
return 0;
}
例题3.5
#include<stdio.h>
#include<math.h>
int main()
{
double a, b, c;
printf("请输入一元二次方程各系数a,b,c,求解");
scanf("%lf%lf%lf", &a, &b, &c);
double der = sqrt(b * b - 4*a*c);
double x1 = (-b + der) / 2 * a,
x2 = (-b - der) / 2 * a;
printf("x1=%7.2f\nx2=%7.2f", x1, x2);
return 0;
}
例题3.6
#include<stdio.h>
int main()
{
double a = 1;
printf("%f\n", a / 3);
return 0;
}
例题3.7
#include<stdio.h>
int main()
{
float a;
a = 10000 / 3.0;
printf("%f\n", a);
return 0;
}
例题3.9
#include<stdio.h>
int main()
{
char a, b, c;
a = getchar();
b = getchar();
c = getchar();
putchar(a);
putchar(b);
putchar(c);
return 0;
}
例题3.10
#include<stdio.h>
int main()
{
char c1, c2;
c1 = getchar();
c2 = c1 + 32;
putchar(c2);
putchar('.\n');
return 0;
}
例题3.8
#include<stdio.h>
int main()
{
char a = 'B', b = 'O', c = 'Y';
putchar(a);
putchar(b);
putchar(c);
return 0;
}
韩信点兵
#include<stdio.h>
#include<math.h>
int main()
{
int n = 0;
do {
n = n + 1;
}
while (n%5!=1||n%6!=5||n%7!=4||n%11!=10);
printf("韩信至少拥有的士兵数为%d",n);
return 0;
}
实验四
#include<stdio.h>
int main()
{
char a, b, c;
scanf("%c %c %c", &a, &b, &c);
int num1=0, num2=0, num3=0;
num1 = int(a - '\0');
num2 = int(b - '\0');
num3 = int(c - '\0');
printf("平均值为%d", (num1 + num2 + num3) / 3);
return 0;
}
c语言判断是否为闰年的三种方法
第一种
#include<stdio.h>
int main()
{
int year, leap;
printf("输入年份：");
scanf("%d", &year);
if (year % 4 == 0)
{
if (year % 100 == 0)
{
if (year % 400 == 0)
leap = 1;
else leap = 0;
}
else leap = 1;
}
else leap = 0;
if (leap) printf("%d是闰年.\n", year);
else printf("%d不是闰年.\n", year);
return 0;
}
第二种
#include<stdio.h>
int main()
{
int x, y, z;
printf("请输入一个年份");
scanf("%d", &x);
z = x % 4;
if (0 == x % 100) z = z + 1;
y = x % 400;
switch (y)
{
case 0:printf("%d是闰年", x); break;
default:
{
switch (z)
{
case 0:printf("%d是闰年", x); break;
case 1:printf("%d不是闰年", x); break;
default:printf("%d不是闰年", x); break;
}
}
}
}
第三种
#include<stdio.h>
int main()
{
int year, leap;
printf("输入一个年份");
scanf("%d", &year);
if (year % 4 == 0 && year % 100 != 0 || (year % 400 == 0))
leap = 1;
else
leap = 0;
if (leap);
printf("%d is leap year", year);
printf("%d is not leap year", year);
return 0; }
第四章例题作业
4.1
#include<stdio.h>
#include<math.h>
int main()
{
double a = 0, b = 0, c = 0, disc = 0, x1, x2;
printf("请分别输入方程的二次系数一次系数与常数，需要回车\n");
scanf("%lf %lf %lf",&a,&b,&c);
disc = pow(b, 2) - 4 * a * c;
if (disc < 0)
{
printf("此方程无解\n");
}
else
{
x1 = ((-b / (2 * a)) + (sqrt(disc) / (2 * a)));
x2 = ((-b / (2 * a)) - (sqrt(disc) / (2 * a)));
printf("方程的解为%7.2f,%7.2f", x1, x2);
}
return 0;
}
4.2
#include<stdio.h>
Int main()
{
float a,b,c;
scanf (“%f,%f”,&a,&b);
if(a>b)
{
c=a;
a=b,
b=c;
}
printf(“%5.2f,%5.2f\n”,a,b);
return 0;
}
4.3
#include<stdio.h>
int main()
{
float a, b, c, d;
scanf("%f,%f,%f", &a, &b, &c);
if (a > b)
{
d = a, a = b, b = d;
}
if (a > c)
{
d = a, a = c, c = d;
}
if (b > c)
{
d = b, b = c, c = d;
}
printf("%5.2f,%5.2f,%5.2f", a, b, c);
return 0;
}
4.4
#include<stdio.h>
int main()
{
char ch;
scanf(" % c", &ch);
ch = (ch >= 'A' && ch <= 'Z') ? (ch + 32) : ch;
printf("%c\n", ch);
return 0;
}
4.5
#include<stdio.h>
int main()
{
int x, y;
scanf("%d", &x);
if (x >= 0)
if (x > 0)y = 1;
else y = 0;
else y = -1;
printf("x=%d,y=%d\n", x, y);
return 0;
}
4.6
#include<stdio.h>
int main()
{
char grade;
scanf("%c", &grade);
printf("你的分数为：");
switch (grade)
{
case'A':printf("85~100\n"); break;
case'B':printf("70~84\n"); break;
case'C':printf("60~69\n"); break;
case'D':printf("<60\n"); break;
default:printf("输错啦！！!\n"); break;
}
return 0;
}
4.8
#include<stdio.h>
int main()
{
int year, leap;
printf("输入年份：");
scanf("%d", &year);
if (year % 4 == 0)
{
if (year % 100 == 0)
{
if (year % 400 == 0)
leap = 1;
else leap = 0;
}
else leap = 1;
}
else leap = 0;
if (leap) printf("%d是闰年.\n", year);
else printf("%d不是闰年.\n", year);
4.9
#include<stdio.h>
#include<math.h>
int main() {
int a, b, c;
printf("请输入:");
scanf("%d%d%d", &a, &b, &c);
double delta, x1, x2, m, n;
delta = b * b - 4 * a * c;
if (delta > 0) {
x1 = (-b + sqrt(delta)) / (2 * a);
x2 = (-b - sqrt(delta)) / (2 * a);
printf("原方程的的解为x1=%f,x2=%f\n", x1, x2);
}
if (delta == 0) {
x1 = (-b) / (2 * a);
x2 = x1;
printf("原方程的的解为x1=%f,x2=%f\n", x1, x2);
}
if (delta < 0) {
m = (-b) / (2 * a);
n = sqrt(-delta) / (2 * a);
printf("原方程的的解为:x1=%8.4f+%8.4fi ; x2=%8.4f-%8.4fi\n ", m, n, m, n);
}
return 0;
}
4.10
#include<stdio.h>
int main()
{
int c, s;
float p, w, d, f;
printf("请输入单价，重量以及距离\n");
scanf("%f,%f,%d", &p, &w, &s);
if (s >= 3000)c = 12;
else c = s / 250;
switch (c)
{
case 0:d = 0; break;
case 1:d = 2; break;
case 2:
case 3:d = 5; break;
case 4:
case 5:
case 6:
case 7:d = 8; break;
case 8:
case 9:
case 10:
case 11:d = 10; break;
case 12:d = 15; break;
}
f = p * w * s * (1 - d / 100);
printf("总运费为%10.2f\n", f);
return 0;
}
第四章章末
第四题
#include<stdio.h>
int main()
{
int a, b, c, x, y, d;
printf("请分别输入三个整数");
scanf("%d %d %d", &a, &b, &c);
x = max(a, b);
y = max(b, c);
d = max(x, y);
printf("最大为%d", d);
return 0;
}
int max(int x, int y)
{
int m;
if (x >= y) m = x;
else m = y;
return m;
}
第五题
#include<stdio.h>
#include<math.h>
int main()
{
double x = 0;
int y = 0, i = 0;
for(i > 0;;)
{
printf("请输入一个小于1000的整数:");
scanf("%lf", &x);
if (x > 1000)
{
printf("请重新输入");
}
else
{
y = sqrt(x);
printf("输入的数字开方为%d", y);
return 0;
}
}
return 0;
}
第六题
#include<stdio.h>
int main()
{
double x = 0, y = 0;
int i;
printf("请输入x的值：\n");
scanf("%lf", &x);
if (x < 1)i = 0;
else if (x >= 1 && x < 10)i = 1;
else if (x >= 10)i = 2;
switch (i)
{
case 0:printf("y=%5.2lf", y = (x)); break;
case 1:printf("y=%5.2lf", y = (2 * x - 1)); break;
case 2:printf("y=%5.2lf", y = (3 * x - 11)); break;
}
return 0;
}
第八题
#include<stdio.h>
int main()
{
int x = 0, i = 0;
printf("请输入你的分数");
scanf("%d", &x);
if (x >= 90 && x <= 100)i = 1;
else if (x >= 80 && x <= 89)i = 2;
else if (x >= 70 && x <= 79)i = 3;
else if (x >= 60 && x <= 69)i = 4;
switch (i)
{
case 0:printf("等级为E"); break;
case 1:printf("等级为A"); break;
case 2:printf("等级为B"); break;
case 3:printf("等级为C"); break;
case 4:printf("等级为D"); break;
}
return 0;
}
第九题
#include<stdio.h>
#include<math.h>
int main()
{
int x = 0, y = 0, z = 0, t = 0, m = 0, i = 0;
printf("请输出最高为五位的数字：");
scanf("%d", &x);
z = x;
t = x;
do
{
z = z / 10;
i++;
} while (z != 0);
switch (i)
{
case 1:printf("为个位数\n");
break;
case 2:printf("为十位数\n");
break;
case 3:printf("为百位数\n");
break;
case 4:printf("为千位数\n");
break;
case 5:printf("为万位数\n");
break;
default:
printf("你小子是不是输入了大于五位的数？\n"); //算位数
return 0;
}
printf("输入的数字逐个输出为:\n");
do
{
i--;
m = t / pow(10, i);
printf("%d\n", m);
t = t - m * pow(10, i);
} while (t != 0);
printf("这个数字逆序输出为：");
do
{
y = x % 10;
printf("%d", y);
x = x / 10;
} while (x != 0); //逆序输出
return 0;
}
第十题
第一种方法
# include <stdio.h>
int main()
{
int i, branch;
double bonus, bon1, bon2, bon4, bon6, bon10;
bon1 = 100000 * 0.1;
bon2 = bon1 + 100000 * 0.075;
bon4 = bon2 + 100000 * 0.05;
bon6 = bon4 + 100000 * 0.03;
bon10 = bon6 + 400000 * 0.015;
printf("请输入当月利润i：");
scanf("%d", &i);
printf("i=%d\n", i);
branch = i / 100000;
if (branch > 10) {
branch = 10;
}
switch (branch) {
case 0:bonus = i * 0.1; break;
case 1:bonus = bon1 + (i - 100000) * 0.075; break;
case 2:
case 3:bonus = bon2 + (i - 200000) * 0.05; break;
case 4:
case 5:bonus = bon4 + (i - 400000) * 0.03; break;
case 6:
case 7:
case 8:
case 9:bonus = bon6 + (i - 600000) * 0.015; break;
case 10:bonus = bon10 + (i - 1000000) * 0.01;
}
printf("奖金是%10.2f\n", bonus);
return 0;
}
第二种方法
#include <stdio.h>
int main()
{
double I, salary = 0;
printf("enter performance:");
scanf_s("%lf", &I);
if (I < 0) {
printf("请输入一个正数\n");
system("pause");
return -1;
}
double salary1 = 100000 * 0.1;//10万的奖金
double salary2 = (200000 - 100000) * 0.075 + salary1;//20万的奖金
double salary3 = (400000 - 200000) * 0.05 + salary2;//40万的奖金
double salary4 = (600000 - 400000) * 0.03 + salary3;//60万的奖金
double salary5 = (1000000 - 600000) * 0.015 + salary4;//100万的奖金
if (I <= 100000) {
salary = I * 0.1;//小于100000按10%提成
}else if (I > 100000 && I <= 200000) {
salary = salary1 + (I - 100000) * 0.075;//多出10万的按比例计算，加上10w的奖金
}else if (I > 200000 && I <= 400000) {
salary = salary2 + (I - 200000) * 0.05;//多出20万的按比例计算，加上20w的奖金
}else if (I > 400000 && I <= 600000) {
salary = salary3 + (I - 400000) * 0.03;//多出40万的按比例计算，加上40w的奖金
}else if (I > 600000 && I <= 1000000) {
salary = salary4 + (I - 600000) * 0.015;//多出60万的按比例计算，加上60w的奖金
}else if (I > 1000000){
salary = salary5 + (I - 1000000) * 0.01;//多出100万的按比例计算，加上100w的奖金
}
printf("salary:%f\n", salary);
system("pause");
return 0;
}
第十一题
#include <stdio.h>
int main() {
int a, b, c, d, t;
printf("请输入4个整数：");
scanf("%d%d%d%d", &a, &b, &c, &d);
if (a > b) {
t = a;
a = b;
b = t;
}
if (a > c) {
t = a;
a = c;
c = t;
}
if (a > d) {
t = a;
a = d;
d = t;
}
if (b > c) {
t = b;
b = c;
c = t;
}
if (b > d) {
t = b;
b = d;
d = t;
}
if (c > d) {
t = c;
c = d;
d = t;
}
printf("将这四个数进行从小到大排序：%d,%d,%d,%d", a, b, c, d);
return 0;
}
第十二题
#include<stdio.h>
#include<math.h>
int main()
{
double x, y;
int i = 0;
printf("请先后输入坐标的（x,y），中间用空格隔开:");
scanf("%lf %lf", &x, &y);
if (
(pow((x - 2), 2) - pow((y - 2), 2)) <= 1 ||
(pow((x + 2), 2) - pow((y + 2), 2)) <= 1 ||
(pow((x + 2), 2) - pow((y - 2), 2)) <= 1 ||
(pow((x - 2), 2) - pow((y - 2), 2)) <= 1
)i = 1;
switch (i)
{
case 1:printf("该点高度为10m"); break;
case 0:printf("该点高度为0m"); break;
}
return 0;
}
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