White algorithm - Simple Plan

Recent stay home, has been playing with the city skyline, and then find how to control population education constitute a problem.

Question A:

对于给定的人口数据N(2^31-1>N>9)，求出应该建立多少个A(小学),B(中学),C(大学)才能满足给定的比例(人口文化教育构成)
已知A单个容量为：300
已知B单个容量为：1000
已知C单个容量为：4500

Conditional A:

Example:

输入：13520
输出：27,5,1

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Note: only finished primary school before they can read, before they can finish high school to university.

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With code:

#include <iostream>
#include <stdio.h>
#include <math.h>

int main()
{
	int N,temp,a=0,b=0,c=0;
	float A, B, C;
	scanf_s("%d", &N);
	temp = floor(N / 10);
	for (a = 0; temp > a * 50; a++);
	for (b = 0; temp * 3 > b * 1000; b++);
	for (c = 0; temp > c * 4500; c++);
	printf("%d,%d,%d", a,b,c);
	return 0;
}

But in fact this is not consistent with the laws of the game, a 1w + urban population, to build more than 20 primary school? Budget is absolutely not enough, completing construction of what it once was, so we introduced the following conditions:

Conditional B:

Building Type	Construction costs	Maintenance costs	capacity
primary school	$10000	$ 160 / week	300 people
High school	$24000	$ 560 / week	1000
the University	$75000	$ 1920 / week	4500 people

Problem B:

对于给定的人口数据N(2^31-1>N>9),一定时间T(T>3周)，求出在T时间内应该建立多少个A(小学),B(中学),C(大学)才能满足给定的比例且花费最少(人口文化教育构成)。
注：学生的课程周期为一周，且不计建筑时间。

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To be continued ...