利用笨办法求解

解题思路:

注意事项:

参考代码:

#include <iostream>

using namespace std;

int main()

{

int a;

int b;

while (cin >> 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 > 20000 && a <= 400000)

{

b = 100000 * 0.1 + 100000 * 0.075 + (a - 200000) * 0.05;

}

else if (a > 400000 && a <= 600000)

{

b = 100000 * 0.1 + 100000 * 0.075 + 200000 * 0.05 + (a - 400000) * 0.03;

}

else if (a > 600000 && a <= 1000000)

{

b = 100000 * 0.1 + 100000 * 0.075 + 200000 * 0.05 + 200000 * 0.03 + (a - 600000) * 0.015;

}

else

{

b = 100000 * 0.1 + 100000 * 0.075 + 200000 * 0.05 + 200000 * 0.03 + 400000 * 0.015+(a-1000000)*0.01;

}

cout << b << endl;

}

return 0;

}