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:
1 2 4 -1
4
7 6 -2 -3
Sample Output:
43
分析：
很简单的模拟题，两组数有正有负，问乘积和最大是多少。
注意两点：
1.优惠券可以不全用完
2.数据比较大，用longlong存储数据，scanf和printf输入/打印。
code：
//来自牛客网一位老哥，代码很简洁，原文cin会超时
#include
#include
using namespace std;
bool cmp(long long a,long long b){
return a>b;
}
long long a[100010],b[100010],sum=0;
int main(){
int x,y;
scanf("%d",&x);
for(int i=0;i0&&b[k]>0){ //这里不用过多限定，因为输入本身就有一定的限定
sum+=a[j]*b[k];
j++;k++;
}
j=x-1;k=y-1;
while(a[j]<0&&b[k]<0){
sum+=a[j]*b[k];
num++;
j--;k--;
}
printf("%lld",sum);
return 0;
}
柳婼code：
//用vector动态数组，不会有溢出危险
#include
#include
#include
using namespace std;
int main() {
int m, n, ans = 0, p = 0, q = 0;
scanf("%d", &m);
vector v1(m);
for(int i = 0; i < m; i++)
scanf("%d", &v1[i]);
scanf("%d", &n);
vector v2(n);
for(int i = 0; i < n; i++)
scanf("%d", &v2[i]);
sort(v1.begin(), v1.end());
sort(v2.begin(), v2.end());
while(p < m && q < n && v1[p] < 0 && v2[q] < 0) {
ans += v1[p] * v2[q];
p++; q++;
}
p = m - 1, q = n - 1;
while(p >= 0 && q >= 0 && v1[p] > 0 && v2[q] > 0) {
ans += v1[p] * v2[q];
p--; q--;
}
printf("%d", ans);
return 0;
}