PAT 1037 Magic Coupon（贪心）

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

总结：这次两道题目都挺简单的，没啥难度，不知道为什么还有一个测试点没有通过（还未找到，未设置重复使用数字的情况）

思路：排完序后，找符号相同的相乘

因为题目要求我们求出最大值，所以最大的正数和最大的正数相乘，依次接下去，直到遇到一正一负的情况结束循环，最小的负数和最小的负数相乘（可以到达最大的正数），依次接下去，直到遇到一正一负的情况

使用vector数组a,b存储两组数 ，sort排序后开始遍历循环

代码：

#include 
#include 
#include 
using namespace std;
 
int main(){
    int n,t;
    long long sum=0;
    vector<int> a,b;
    cin >> n;
    
    for(int i=0;i
        scanf("%d",&t);
        a.push_back(t);
    }
    sort(a.begin(),a.end());
    
    cin >> n;
    for(int i=0;i
        scanf("%d",&t);
        b.push_back(t);
    }
    sort(b.begin(),b.end());
    
    bool sign=false;
    for(int i=0,j=0;isize() && jsize();i++,j++){
        if(a[i]*b[i]<0){
            sign=true;
            break;
        }
        sum+=a[i]*b[j];
    }
 
    if(sign){
        for(int i=a.size()-1,j=b.size()-1;i>=0 && j>=0;i--,j--){
            if(a[i]*b[i]<0)     break;
            sum+=a[i]*b[i];
        }
    }
    printf("%lld\n",sum);
    
    return 0;
}

大佬的代码：

思路：排序完后，先找都是负数的相乘，再找都是正数的相乘

#include 
#include 
#include 
using namespace std;
int main() {
    int m, n, ans = 0, p = 0, q = 0;
    scanf("%d", &m);
    vector<int> v1(m);
    for(int i = 0; i < m; i++)
        scanf("%d", &v1[i]);
    scanf("%d", &n);
    vector<int> 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;
}

好好学习，天天向上！

我要考研！

2022.11.4

/*
    思路：这道题目实质上就是匹配数字，将数字匹配然后相乘（且每个值数字只能使用一次），结果最大
    先排序，因为是从小到大排序的，所以需要先从尾巴开始循环，这样能先使用最大的值相乘，结果最大
    如果中间两个数字符号不一样（a[i]*b[i]<0），则结束循环，然后再从头开始循环，未使用过的且
    符号相同
*/
#include 
#include 
using namespace std;
 
int main(){
    int n,m;
    scanf("%d",&n);
    int a[n];
    for(int i=0;iscanf("%d",&a[i]);
    scanf("%d",&m);
    int b[m];
    for(int i=0;iscanf("%d",&b[i]);
    
    int sum=0;
    sort(a,a+n);
    sort(b,b+m);
    int i,j;
    for(i=n-1,j=m-1;i>=0 && j>=0 && a[i]*b[j]>0;i--,j--)
        sum+=a[i]*b[j];
    int q=i,t=j;
    for(i=0,j=0;i0;i++,j++)
        sum+=a[i]*b[j];
    
    printf("%d\n",sum);
    return 0;
}