SRM 480 Cryptography

Problem Statement TopCoder Security Agency (TSA, established today) has just invented a new encryption system! This encryption system takes as its input a list of numbers to encrypt. You work at TSA and your task is to implement a very important part of the encryption process. You are allowed to pick one number in the input list and increment its value by 1. This should be done in such way that the product of all numbers in the list after this change becomes as large as possible. Given the list of numbers as vector numbers, return the maximum product you can obtain. It is guaranteed that the return value will not exceed 2^62. Definition Class: Cryptography Method:encrypt Parameters:vector Returns:long long Method signature:long long encrypt(vector numbers) (be sure your method is public) Constraints – numbers will contain between 2 and 50 elements, inclusive. – Each element of numbers will be between 1 and 1000, inclusive. – The return value will not exceed 2^62. Examples 0) {1,2,3} Returns: 12 If we increment the first number, we get 2*2*3 = 12. If we increment the second, we get 1*3*3 = 9. If we increment the third, we get 1*2*4 = 8. Hence, the correct return value is 12. 1) {1,3,2,1,1,3} Returns: 36 The elements of numbers are not necessarily unique. 2) {1000,999,998,997,996,995} Returns: 986074810223904000 The answer may be very big, but will not exceed 2^62. 3) {1,1,1,1} Returns: 2 This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited.

#include <vector>
using namespace std;

class Cryptography {
    public:
        long long encrypt(vector <int> numbers) {
            long long ans = 0;

            for(int i=0; i<numbers.size(); i++) {
                long long seki=1;
                for (int j=0; j<numbers.length; j++){
                    if (i==j) {
                        seki *= (numbers[j] + 1);
                    } else {
                        seki *= numbers[j];
                    }
                }
                ans = max(ans, seki);
            }
            return ans;
        }
}

この問題は小さい数字に1を足す。その場合、増加率が最も大きくなる。

#include <vector>
#include <algorithm>
using namespace std;

class Cryptography {
    public:
        long long encrypt(vector <int> numbers) {
            long long ret = 1;
            sort(numbers.bigin(), numbers.end());
            numbers[0]++;

            for (int i = 0; i < number.size(); i++){
                ret *= numbers[i];
            }
            return ret;
        }
}