1037 Magic Coupon(简单题，sort()排序)

目录

题目

测试样例

输入样例

输出样例

提交结果截图

带详细注释的源代码

题目

题目链接：

1037 Magic Coupon https://pintia.cn/problem-sets/994805342720868352/problems/994805451374313472

测试样例

输入样例

4
1 2 4 -1
4
7 6 -2 -3

输出样例

43

提交结果截图

带详细注释的源代码

#include <cstdio>
#include "string.h"
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

//正数从小到大排序
bool cmp1(const int a, const int b)
{
	return a > b;
}

//负数从大到小排序
bool cmp2(const int a, const int b)
{
	return a < b;
}

int main()
{
	vector<int>c_p, c_n, p_p, p_n;//coupon_positive, coupon_negetive, product_positive, product_negetive
	int m, n, tmp, sum = 0;
	cin >> m;
	for (int i = 0; i < m; i++)
	{
		cin >> tmp;
		if (tmp > 0)//不能用if(tmp)，因为只有0为假，其他都为真
			c_p.push_back(tmp);
		else
			c_n.push_back(tmp);
	}
	cin >> n;
	for (int i = 0; i < n; i++)
	{
		cin >> tmp;
		if (tmp > 0)
			p_p.push_back(tmp);
		else
			p_n.push_back(tmp);
	}
	//排序
	sort(c_p.begin(), c_p.end(), cmp1);
	sort(c_n.begin(), c_n.end(), cmp2);
	sort(p_p.begin(), p_p.end(), cmp1);
	sort(p_n.begin(), p_n.end(), cmp2);
	//先求正乘正
	for (int i = 0, j = 0; i < c_p.size() && j < p_p.size();i++, j++)
		sum += c_p[i] * p_p[j];//cout<<"c_p["<<i<<"]"<< c_p[i]<<", p_p["<<j<<"] = "<< p_p[j]<<endl;

	//再求负乘负
	for (int i = 0, j = 0; i < c_n.size() && j < p_n.size();i++, j++)
		sum += c_n[i] * p_n[j];// cout << "c_n[" << i << "]" << c_n[i] << ", p_n[" << j << "] = " << p_n[j] << endl;
	cout << sum << endl;
	return 0;
}

标签：sort,tmp,begin,Magic,Coupon,样例,int,include
来源： https://blog.csdn.net/NOT_TODAY1/article/details/120614776