

#include <stdio.h>
#include <math.h>
int main()
{
float a,b,c;
float delta,realpart,imagpart,x1,x2;
printf("请输入一元二次方程的系数:");
scanf("%f,%f,%f",&a,&b,&c);
if (a==0){
printf("错误\n");
return 1;
}
delta=b*b-4*a*c;
if (delta>0){
x1=(-b+sqrt(delta))/(2*a);
x2=(-b-sqrt(delta))/(2*a);
printf("方程有两个不同的实根:\n");
printf("x1=%f\n",x1);
printf("x2=%f\n",x2);
}
else if(delta==0){
x1=-b/(2*a);
printf("方程只有一个实根:%f",x1);
}
else{
realpart=-b/(2*a);
imagpart=delta/(2*a);
x1=realpart+imagpart;
x2=realpart-imagpart;
printf("方程有两个虚根:\nx1=%f\nx2=%f",x1,x2);
}
return 0;
}


#include <stdio.h>
#include <math.h>
int main()
{
float a,b,c;
printf("请输入三角形的三边长:");
scanf("%f,%f,%f",&a,&b,&c);
if(a+b<=c||a+c<=b||b+c<=a){
printf("不能构成三角形");
return 1;
}
else if(a==b&&b==c){
printf("等边三角形\n");
}
else if(a==b||b==c||a==c){
if(a*a+b*b==c*c||a*a+c*c==b*b||b*b+c*c==a*a){
printf("等腰直角三角形");
}else{
printf("等腰三角形");
}
}else if(a*a+b*b==c*c||a*a+c*c==b*b||b*b+c*c==a*a){
printf("直角三角形");
}else{
printf("一般三角形");
}
return 0;
}


#include <stdio.h>
#include <math.h>
int main()
{
int data1,data2;
char op;
printf("please enter an expression:");
scanf("%d%c%d",&data1,&op,&data2);
switch(op)
{
case'+':
printf("%d+%d=%d\n",data1,data2,data1+data2);
break;
case'-':
printf("%d-%d=%d\n",data1,data2,data1-data2);
break;
case'*':
printf("%d*%d=%d\n",data1,data2,data1*data2);
break;
case'/':
if (data2==0)
printf("错误\n");
else
printf("%d/%d=%d\n",data1,data2,data1/data2);
break;
default:
printf("invalid operator\n");
}
return 0;
}

#include <stdio.h>
int main() {
int year, month, day;
int days_in_month[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int day_of_year = 0;
printf("请输入年份、月份、日期(格式:年 月 日,用空格分隔:)");
scanf("%d %d %d", &year, &month, &day);
int is_leap_year = 0;
if ((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0)) {
is_leap_year = 1;
}
if (is_leap_year) {
days_in_month[1] = 29;
}
if (month < 1 || month > 12) {
printf("月份输入不合法!\n");
return 1;
}
if (day < 1 || day > days_in_month[month - 1]) {
printf("日期输入不合法!\n");
return 1;
}
for (int i = 0; i < month - 1; i++) {
day_of_year += days_in_month[i];
}
day_of_year += day;
printf("%d年%d月%d日 是 %d 年的第 %d 天\n", year, month, day, year, day_of_year);
return 0;
}

#include <stdio.h>
#include <math.h>
int main() {
int region_code;
float weight, total_fee;
printf("请输入目的区域编码 (0-4): ");
scanf("%d", ®ion_code);
printf("请输入邮件重量 (kg): ");
scanf("%f", &weight);
if (region_code < 0 || region_code > 4 || weight <= 0) {
printf("无效的输入,请检查区域编码和重量。\n");
return 1;
}
float base_fee, extra_fee_per_kg;
if (region_code == 0) {
base_fee = 10;
extra_fee_per_kg = 3;
} else if (region_code == 1) {
base_fee = 10;
extra_fee_per_kg = 4;}
else {
base_fee = 15;
if (region_code == 2) {
extra_fee_per_kg = 5;
} else if (region_code == 3) {
extra_fee_per_kg = 6.5;
} else { // region_code == 4
extra_fee_per_kg = 10;
}
}
float extra_weight = weight - 1;
if (extra_weight > 0) {
extra_weight = ceil(extra_weight); // 不足1kg按1kg计算
} else {
extra_weight = 0; // 重量 ≤1kg,无续重
}
total_fee = base_fee + extra_weight * extra_fee_per_kg;
printf("总运费为: %.2f 元\n", total_fee);
return 0;
}

#include <stdio.h>
int main() {
double profit, bonus = 0.0;
printf("请输入当月利润(单位:万元):");
scanf("%lf", &profit);
if (profit <= 10) {
bonus = profit * 0.1;
} else if (profit <= 20) {
bonus = 10 * 0.1 + (profit - 10) * 0.075;
} else if (profit <= 40) {
bonus = 10 * 0.1 + 10 * 0.075 + (profit - 20) * 0.05;
} else if (profit <= 60) {
bonus = 10 * 0.1 + 10 * 0.075 + 20 * 0.05 + (profit - 40) * 0.03;
} else if (profit <= 100) {
bonus = 10 * 0.1 + 10 * 0.075 + 20 * 0.05 + 20 * 0.03 + (profit - 60) * 0.015;
} else {
bonus = 10 * 0.1 + 10 * 0.075 + 20 * 0.05 + 20 * 0.03 + 40 * 0.015 + (profit - 100) * 0.01;
}
printf("应发奖金总数为:%.2f 万元\n", bonus);
return 0;
}

#include <stdio.h>
int main() {
double profit, bonus = 0.0;
int range;
printf("请输入当月利润(单位:万元):");
scanf("%lf", &profit);
if (profit <= 10) {
range = 0;
} else if (profit <= 20) {
range = 1;
} else if (profit <= 40) {
range = 2;
} else if (profit <= 60) {
range = 3;
} else if (profit <= 100) {
range = 4;
} else {
range = 5;
}
switch(range) {
case 0:
bonus = profit * 0.1;
break;
case 1:
bonus = 10 * 0.1 + (profit - 10) * 0.075;
break;
case 2:
bonus = 10 * 0.1 + 10 * 0.075 + (profit - 20) * 0.05;
break;
case 3:
bonus = 10 * 0.1 + 10 * 0.075 + 20 * 0.05 + (profit - 40) * 0.03;
break;
case 4:
bonus = 10 * 0.1 + 10 * 0.075 + 20 * 0.05 + 20 * 0.03 + (profit - 60) * 0.015;
break;
case 5:
bonus = 10 * 0.1 + 10 * 0.075 + 20 * 0.05 + 20 * 0.03 + 40 * 0.015 + (profit - 100) * 0.01;
break;
default:
printf("无效的利润区间。\n");
return 1;
}
printf("应发奖金总数为:%.2f 万元\n", bonus);
return 0;
}
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