目录
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;
}