关于#c语言#的问题，请各位专家解答！

Question

搞！！！搞几遍老错，老师让我们用的vs**

1.

Answer 1

//if写法
#include<stdio.h>
int main() {
    double a,b;//a表示利润,b表示奖金提成
    scanf_s("%lf", &a);
    if (a <= 100000)
        b = a * 0.1;
    else if (a > 100000 && a <= 200000)
        b = 100000 * 0.1 + (a - 100000) * 0.075;
    else if (a > 200000 && a <= 400000)
        b = 100000 * (0.1 + 0.075) + (a - 200000) * 0.05;
    else if (a > 400000 && a <= 600000)
        b = 100000 * (0.1 + 0.075) + 200000 * 0.05 + (a - 400000) * 0.03;
    else if (a > 600000 && a <= 1000000)
        b = 100000 * (0.1 + 0.075) + 200000 * (0.05 + 0.03) + (a - 600000) * 0.015;
    else if (a > 1000000)
        b = 100000 * (0.1 + 0.075) + 200000 * (0.05 + 0.03) + 400000 * 0.015 + (a - 1000000) * 0.01;
    printf("%lf", b);
    return 0;
}

//switch写法
#include<stdio.h>
int main()
{
    double a, x1, x2, x4, x6, x10;
    double b;
    int n;
    x1 = 100000 * 0.1;//总额高于10 000时10 000以下的奖励 
    x2 = x1 + 100000 * 0.075;//总额高于20 000时20 000以下的奖励 
    x4 = x2 + 200000 * 0.05;//总额高于40 000时40 000以下的奖励 
    x6 = x4 + 200000 * 0.03;//总额高于60 000时60 000以下的奖励 
    x10 = x6 + 400000 * 0.015;//总额高于1 000 000时1 000 000以下的奖励 
    scanf("%lf", &a);
    n = a / 100000;
    if (n > 10)
        n = 10;
    switch (n)
    {
    case 0: 
        b = a * 0.1; break;
    case 1: 
        b = x1 + (a - 100000) * 0.075; break;
    case 2:    case 3: 
        b = x2 + (a - 200000) * 0.05; break;
    case 4:    case 5: 
        b = x4 + (a - 400000) * 0.03; break;
    case 6:    case 7:    case 8:    case 9:
        b = x6 + (a - 600000) * 0.015; break;
    case 10: 
        b = x10 + (a - 1000000) * 0.01; break;
    }
    printf("%lf", b);
}