Article Directory
1. Title
Give you an array of positive integers nums, you need to choose some subsets from it , then multiply each number in the subset by an arbitrary integer, and find their sum.
If the result of the sum is 1, then the original array is a "good array", return True; otherwise, return False.
示例 1：
输入：nums = [12,5,7,23]
输出：true
解释：挑选数字 5 和 7。
5*3 + 7*(-2) = 1
示例 2：
输入：nums = [29,6,10]
输出：true
解释：挑选数字 29, 6 和 10。
29*1 + 6*(-3) + 10*(-1) = 1
示例 3：
输入：nums = [3,6]
输出：false
提示：
1 <= nums.length <= 10^5
1 <= nums[i] <= 10^9
Source: LeetCode (LeetCode)
Link: https://leetcode-cn.com/problems/check-if-it-is-a-good-array
Copyright is owned by LeetCode . For commercial reprints, please contact the official authorization. For non-commercial reprints, please indicate the source.
2. Problem solving
- Insert numbers into set, traverse backward
- The current number is modulo the previous number, modulo == 1, return true
- mold! = 1, insert the die into the set
class Solution {
public:
bool isGoodArray(vector<int>& nums) {
set<int> s(nums.begin(), nums.end());
if(*s.begin()==1) return true;
auto it = s.rbegin();
int prev = *it, mod;
for(it++; it != s.rend(); ++it)
{
if(*it == 0)
break;
mod = prev%*it;
if(mod == 1)
return true;
else
s.insert(mod);
prev = *it;
}
return false;
}
};
436 ms 44 MB
Look at the answer again: Pei Shu theorem
- Find the greatest common divisor of all numbers, it needs to be 1
class Solution {
public:
bool isGoodArray(vector<int>& nums) {
int g = nums[0];
for(int i = 1; i < nums.size(); ++i)
{
g = __gcd(g, nums[i]);
}
return g == 1;
}
};
class Solution {
public:
bool isGoodArray(vector<int>& nums) {
int g = nums[0];
for(int i = 1; i < nums.size(); ++i)
{
g = gcd(g, nums[i]);
}
return g == 1;
}
int gcd(int a, int b)
{
int r;
while(b)
{
r = a%b;
a = b;
b = r;
}
return a;
}
};
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