程序分析：请利用数轴来分界，定位。注意定义时需把奖金定义成长整型。

#include<stdio.h>
int main()
{
    double i;
    double bonus1,bonus2,bonus4,bonus6,bonus10,bonus;
    printf("你的净利润是：\n");
    scanf("%lf",&i);
    bonus1=100000*0.1;
    bonus2=bonus1+100000*0.075;
    bonus4=bonus2+200000*0.05;
    bonus6=bonus4+200000*0.03;
    bonus10=bonus6+400000*0.015;
    if(i<=100000) {
        bonus=i*0.1;
    } else if(i<=200000) {
        bonus=bonus1+(i-100000)*0.075;
    } else if(i<=400000) {
        bonus=bonus2+(i-200000)*0.05;
    } else if(i<=600000) {
        bonus=bonus4+(i-400000)*0.03;
    } else if(i<=1000000) {
        bonus=bonus6+(i-600000)*0.015;
    } else if(i>1000000) {
        bonus=bonus10+(i-1000000)*0.01;
    }
    printf("提成为：bonus=%lf",bonus);
 
    printf("\n");
}

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);
}

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;
}

利用 switch 的击穿现象

#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;
}