程序设计与算法（一）第11周测验(2019夏季)_程序设计与算法(一)第11周测验-优快云博客

本文探讨了两个有趣的计算问题：一是如何通过二分查找法确定从多个圆形区域中分配特定数量的派的最大可能尺寸；二是如何在限定月份数下找到将开销合理分配至各月的最大值。通过具体的算法实现，展示了二分查找法在解决此类问题中的应用。

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001:派

#include <iostream>
#include <iomanip>
#include<cstring>
#pragma warning(disable:4996)
using namespace std;
const double EPS = 1.0E-5;
const double PI = 3.141592653589793;
const int NUM = 10000;
double pie[NUM];
double testpie[NUM];
bool feasible(double size, double pie[], int n,int f) {
	int sum = 0;
	for (int i = 0;i < n;i++)
		sum += int(pie[i] / size);
	if (sum >= f)
		return true;
	else
		return false;
}
double BinarySearch(double pie[],double limit,int n,int f) {
	double l = 0, r=limit,lastSize=0;
	while (r - l > EPS) {
		double mid = (r + l) / 2;
		if (feasible(mid, pie, n,f)) {
			lastSize = mid;
			l = mid;
		}
		else
			r = mid;
	}
	return lastSize;
}

int main()
{
	//freopen("C:\\Users\\czh\\Desktop\\2.txt", "r", stdin);
	int n, f;
	cin >> n >> f;
	f++;
	double limit=0, sum = 0;
	for (int i = 0;i < n;i++) {
		cin >> pie[i];
		pie[i] =pie[i] * pie[i] * PI;
		sum += pie[i];
		limit = limit < pie[i] ? pie[i] : limit;
	}
	limit = limit > sum / f ? sum / f : limit;
	cout << fixed << setprecision(3) << BinarySearch(pie, limit, n,f);
	//printf("%.3f", BinarySearch(pie, limit, n, f));
	return 0;
}

002:月度开销

#include<iostream>
#include<cstdlib>
#pragma warning(disable:4996)
using namespace std;
const int NUM = 100000;
int Expense[NUM];
bool IsOk(int n,int m,int Max) {
	int sum = 0, count = 1;
	for (int i = 0;i < n;i++) {
		if (sum + Expense[i] > Max) {
			sum = Expense[i];
			count++;
		}
		else
			sum += Expense[i];
	}
	if (count <= m)return true;
	else return false;
}
int BinarySearch(int n, int m,int Max,int sum) {
	int l = Max, r = sum, lastExp = -1;
	while (l <= r) {
		int mid = (l + r) / 2;
		if (IsOk(n, m, mid)) {
			lastExp = mid;
			r = mid - 1;
		}
		else
			l = mid + 1;
	}
	return lastExp;
}
int main()
{
	//freopen("C:\\Users\\czh\\Desktop\\2.txt", "r", stdin);
	int n, m, Max =0,sum=0;
	cin >> n >> m;
	for (int i = 0;i < n;i++) {
		cin >> Expense[i];
		Max = Expense[i] > Max ? Expense[i] : Max;
		sum += Expense[i];
	}
	cout << BinarySearch(n, m, Max, sum);
}