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The magic shop in Mars is offering some magic coupons. Each coupon has an integer N printed on it, meaning that when you use this coupon with a product, you may get N times the value of that product back! What is more, the shop also offers some bonus product for free. However, if you apply a coupon with a positive N to this bonus product, you will have to pay the shop N times the value of the bonus product... but hey, magically, they have some coupons with negative N‘s!
For example, given a set of coupons {1 2 4 -1}, and a set of product values {7 6 -2 -3} (in Mars dollars M$) where a negative value corresponds to a bonus product. You can apply coupon 3 (with N being 4) to product 1 (with value M$7) to get M$28 back; coupon 2 to product 2 to get M$12 back; and coupon 4 to product 4 to get M$3 back. On the other hand, if you apply coupon 3 to product 4, you will have to pay M$12 to the shop.
Each coupon and each product may be selected at most once. Your task is to get as much money back as possible.
Input Specification:
Each input file contains one test case. For each case, the first line contains the number of coupons NC, followed by a line with NC coupon integers. Then the next line contains the number of products NP, followed by a line with NP product values. Here 1<= NC, NP <= 105, and it is guaranteed that all the numbers will not exceed 230.
Output Specification:
For each test case, simply print in a line the maximum amount of money you can get back.
Sample Input:4 1 2 4 -1 4 7 6 -2 -3Sample Output:
43
#pragma warning(disable:4996)#include <iostream>#include <vector>#include <algorithm>using namespace std;vector<long long int> Nina0, Nouta0,Ninu0,Noutu0;char result[100] = { ‘0 ‘};bool cmp1(int a, int b) {return a > b;}bool cmp2(int a, int b) {return a < b;}void AddSum(long long int a) {int jinwei = 0;int temp;int i = 0;while (a!=0||jinwei!=0) {temp = a % 10;a = a / 10;if (temp + result[i] + jinwei <= ‘9‘) {result[i] = temp + result[i] + jinwei;jinwei = 0;}else {result[i] = temp + result[i] + jinwei - 10;jinwei = 1;}i++;}}int main(void) {//long long int k = 1073741825;//long long int p;//p = k*k *10000;//cout << p;int nc, np;cin >> nc;for (int i = 0; i < 99; i++)result[i] = ‘0‘;for (int i = 0; i < nc; i++) {long long int temp;cin >> temp;if (temp > 0)Nina0.push_back(temp);elseNinu0.push_back(temp);}cin >> np;for (int i = 0; i < np; i++) {int temp2;cin >> temp2;if (temp2 > 0)Nouta0.push_back(temp2);elseNoutu0.push_back(temp2);}if(Nina0.empty()==false)sort(Nina0.begin(), Nina0.end(), cmp1);if(Nouta0.empty()==false)sort(Nouta0.begin(), Nouta0.end(), cmp1);if (Ninu0.empty() == false)sort(Ninu0.begin(), Ninu0.end(), cmp2);if (Noutu0.empty() == false)sort(Noutu0.begin(), Noutu0.end(), cmp2);int index;long long int sum = 0;if (Nina0.empty() == false && Ninu0.empty() == false && Nouta0.empty() == false && Noutu0.empty() == false) {if (Nina0.size() >= Nouta0.size()) {index = Nouta0.size();}elseindex = Nina0.size();for (int i = 0; i < index; i++) {sum = Nina0[i] * Nouta0[i];AddSum(sum);}if (Ninu0.size() <= Noutu0.size()) {index = Ninu0.size();}elseindex = Noutu0.size();for (int i = 0; i < index; i++) {sum = Ninu0[i] * Noutu0[i];AddSum(sum);}}else if (Nina0.empty() == true || Nouta0.empty() == true) {if (Ninu0.size() <= Noutu0.size()) {index = Ninu0.size();}elseindex = Noutu0.size();for (int i = 0; i < index; i++) {sum = Ninu0[i] * Noutu0[i];AddSum(sum);}}else if (Ninu0.empty() == true || Noutu0.empty() == true) {if (Nina0.size() >= Nouta0.size()) {index = Nouta0.size();}elseindex = Nina0.size();for (int i = 0; i < index; i++) {sum = Nina0[i] * Nouta0[i];AddSum(sum);}}int flag = 0;for (int i = 99; i >= 0; i--) {if (result[i] > ‘0‘) {flag = i;break;}}for (int i = flag; i >= 0; i--)cout << result[i];return 0;}
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原文地址：http://www.cnblogs.com/zzandliz/p/5023131.html
