PAT--1037. Magic Coupon

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 -3

Sample Output:

43

题解

对nc[]和np[] 按照正数, 0, 负数并按绝对值降序的次序排序。
然后两个指针i, j扫描nc和np，符号相同的作乘积累加。

#include <bits/stdc++.h>using namespace std;const int maxn = 100010;int nc, np;int a[maxn], b[maxn];int cmp(int x, int y){    if(x >= 0 && y >= 0) return x > y;    if(x <= 0 && y <= 0) return abs(x) > abs(y);    if(x >= 0 && y <= 0) return 1;    if(x <= 0 && y >= 0) return 0;}bool sign(int x, int y){    return (x > 0 && y > 0) || (x < 0 && y < 0);}int main() {#ifndef ONLINE_JUDGEfreopen("data.in", "r", stdin);#endif // ONLINE_JUDGE    scanf("%d", &nc);    for(int i = 0; i < nc; ++i) scanf("%d", a + i);    scanf("%d", &np);    for(int i = 0; i < np; ++i) scanf("%d", b + i);    sort(a, a + nc, cmp);    sort(b, b + np, cmp);    long long ans = 0;    int i = 0, j = 0;    while(i < nc && j < np){        if(sign(a[i], b[j])) {            ans += a[i] * b[j];            i++, j++;        }else if(a[i] <= 0 && b[j] >= 0) j++;        else i++;    }    printf("%lld\n", ans);    return 0;}