C 练习实例2

#6

14

优化如下:

#include <stdio.h>

#define WAN 10000

int main()
{
    double I = 0; // 利润
    double B = 0; // 奖金
    
    scanf("%lf", &I);
    I /= WAN;

    if (I > 100 * WAN)
    {
        B += ((I - 100) * 0.01);
        I = 100;
    }

    if (I > 60)
    {
        B += ((I - 60) * 0.015);
        I = 60;
    }
    if (I > 40)
    {
        B += ((I - 40) * 0.03);
        I = 40;
    }
    
    if (I > 20)
    {
        B += ((I - 20) * 0.05);
        I = 20;
    }

    if (I > 10)
    {
        B += ((I - 10) * 0.075);
        I = 10;
    }

    B += (I * 0.1);

    printf("%lf", B);
}

DCGDDD3年前 (2017-08-13)

#5

3

尝试使用循环优化代码的适用性：

#include<stdio.h>
int main()
{
    int i;
    double lirun;
    double jiangjin = 0;
    float fanwei[] = {100000, 200000, 400000, 600000, 1000000};
    float ticheng[] = {0.1, 0.075, 0.05, 0.03, 0.015, 0.01};
    printf("您好，请问您的净利润是多少？\n");
    scanf("%lf", &lirun);
    for (i=0;i<5;i++)
    {
        if (lirun < fanwei[i])
        {
            jiangjin += lirun * ticheng[i];
            break;
        }
        else
        {
            jiangjin += fanwei[i] * ticheng[i];
            lirun -= fanwei[i];
        }
    }
    printf("奖金是%.2lf\n", jiangjin);

    return 0;
}

yarzzz2年前 (2018-01-12)

#4

2

利用 switch 的击穿现象，分享个计算方法。

#include <stdio.h>

int main(){
    double d;
    int money = 100000;
    float res=0.0;
    int flag;
    scanf("%lf",&d);
    flag = (int)(d/money);
    flag = flag >10?10:flag;
    switch(flag){
        case 10:
            res += (d-10*money)*0.01;
            d = 10*money;
        case 9:
        case 8:
        case 7:
        case 6:
            res += (d-6*money)*0.015;
            d = 6*money;
        case 5:
        case 4:
            res+= (d-4*money)*0.03;
            d = 4*money;
        case 3:
        case 2:
            res += (d-2*money)*0.05;
            d = 2*money;
        case 1:
            res += (d-money)*0.075;
            d = money;
        case 0:
            res += d *0.1;
    }
    
    printf("%.2f\n",res);
    return 0;
}

言灵之术2年前 (2018-03-10)

#3

1

#include "stdafx.h"


/*
Section 与 Percent 对应关系：
   利润<=Section[i] 时 ， 提成为Percent[i]
*/ 
int Section[] = { 100000, 200000, 400000, 600000, 1000000 };
double Percent[] = { (double)10 / 100, 7.5 / 100, (double)5 / 100, (double)3 / 100, 1.5 / 100 };

//超过最大限定值时的提成比
double OtherPercent = (double)1 / 100;

int Len = sizeof(Section) / sizeof(int);


/*
说明：
  index 在[0,Len) 范围内时，说明 Profit<= Section[Len-1] 或 Profit<=Section[0]
  index = Len 时， 说明 Profit > Section[Len-1]
*/
int Split(const double& profit)
{
    int index = -1, i=0;
    if (profit <= 0.0)
        return index;
    for (; i < Len; i++)
    {
        if ((profit/Section[i]) >1.0)
            continue;

        index = i;
        break;
    }

    //profit>Section[Len-1]时，index赋值为Len
    if (i == Len)
        index = i;
    return index;
}



int _tmain(int argc, _TCHAR* argv[])
{
    double Profit=0.0,  GiveOut=0.0;
    //输入利润
    printf_s("利润：");
    scanf_s("%lf", &Profit);

    

    //计算应发放奖金
    int index = Split(Profit);
    double tempprofit = Profit;
    while (index>=0)
    {
        //超过最大限定值
        if (index == Len)
        {
            GiveOut += (tempprofit - Section[Len-1]) * OtherPercent;
            tempprofit = Section[Len-1];
        }
        else
        {
            double temp;
            //tempprofit 是否在 Section[0] 范围内
            temp = (index-1)>=0? tempprofit-Section[index-1] : tempprofit-0;
            GiveOut += temp * Percent[index];
            tempprofit = (index-1)>=0 ? Section[index-1]:0;
        }
        index = Split(tempprofit);
    }

    printf_s("应发放奖金:%g\n", GiveOut);
    return 0;
}

阿雪2年前 (2018-05-23)

#2

5

对笔记2进行更正:

#include<stdio.h>
int main(){
    int i;
    double lirun;
    double a;
    double jiangjin = 0;
    float ticheng[] = {0,0.1, 0.075, 0.05, 0.03, 0.015, 0.01};    
    printf("您好，请问您的净利润是多少？\n");
    scanf("%lf", &lirun);
    a=lirun;
    float fanwei[] = {0,100000, 200000, 400000, 600000, 1000000, a};
    for (i=1;i<7;i++) {
        if (a < fanwei[i]){
        jiangjin += lirun * ticheng[i];
            break;        
        }       
        else   
        {       
            jiangjin += (fanwei[i]-fanwei[i-1]) * ticheng[i];  
            lirun -= fanwei[i];    
        }
    }  
    printf("奖金是%.2lf\n", jiangjin);

    return 0;
}

Latnetop1年前 (2018-10-02)

#1

1

参考方法：

#include <stdio.h>

#define MAX_ARRAY_PROFIT 6  
const double profit1[MAX_ARRAY_PROFIT]={0,100000,200000,400000,600000,1000000};
const double percent[MAX_ARRAY_PROFIT]={0.1,0.075,0.05,0.03,0.015,0.01};

int main()
{
    unsigned char i;
    double p;
    double bonus;
    printf("你的净利润是：\n");
    scanf("%lf",&p);

    for(i=0,bonus=0;i<MAX_ARRAY_PROFIT-1;i++)
    {
        if(p>profit1[i] && p<profit1[i+1]) {
            bonus +=(p-profit1[i])*percent[i];
        }

        if(p>=profit1[i+1]){
            bonus +=(profit1[i+1]-profit1[i])*percent[i];
        }
    }

    if(p>=profit1[MAX_ARRAY_PROFIT-1]){
        bonus +=(p-profit1[MAX_ARRAY_PROFIT-1])*percent[MAX_ARRAY_PROFIT-1];
    }
    printf("提成为：bonus=%lf",bonus);
    printf("\n");
    return 0;
}

ALLEN1年前 (2018-10-17)

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