给定两个正整数 x 和 y,如果某一整数等于 xi + yj,其中整数 i >= 0 且 j >= 0,那么我们认为该整数是一个强整数。
返回值小于或等于 bound 的所有强整数组成的列表。
你可以按任何顺序返回答案。在你的回答中,每个值最多出现一次。
示例 1:
输入:x = 2, y = 3, bound = 10
输出:[2,3,4,5,7,9,10]
解释: 2 = 20 + 30 3 = 21 + 30 4 = 20 + 31 5 = 21 + 31 7 = 22 + 31 9 = 23 + 30 10 = 20 + 32
示例 2:
输入:x = 3, y = 5, bound = 15
输出:[2,4,6,8,10,14]
提示:
1 <= x <= 100
1 <= y <= 100
0 <= bound <= 10^6
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/powerful-integers 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
class Solution {
public:
vector<int> powerfulIntegers(int x, int y, int bound) {
int i, j, res;
unordered_set<int> s;
int Maxi = (x != 1) ? log(bound)/log(x)+1 : bound/2;
int Maxj = (y != 1) ? log(bound)/log(y)+1 : bound/2;
for(i = 0; i <= Maxi; i++)
{
for(j = 0; j <= Maxj; j++)
{
res = pow(x,i)+pow(y,j);
if(res <= bound)
s.insert(res);
}
}
return vector<int> (s.begin(),s.end());
}
};
class Solution {
public:
vector<int> powerfulIntegers(int x, int y, int bound) {
int i = 0, j = 0, res;
unordered_set<int> s;
int powx, powy;
for(i = 0; (powx = pow(x,i)) <= bound; i++)
{
for(j = 0; (powy = pow(y,j)) <= bound; j++)
{
res = powx+powy;
if(res <= bound)
s.insert(res);
if(y == 1)
break;
}
if(x == 1)
break;
}
return vector<int> (s.begin(),s.end());
}
};
or
class Solution {
public:
vector<int> powerfulIntegers(int x, int y, int bound) {
int i, j, res;
unordered_set<int> s;
int Maxi = (x != 1) ? log(bound)/log(x)+1 : 1;
int Maxj = (y != 1) ? log(bound)/log(y)+1 : 1;
for(i = 0; i < Maxi; i++)
{
for(j = 0; j < Maxj; j++)
{
res = pow(x,i)+pow(y,j);
if(res <= bound)
s.insert(res);
}
}
return vector<int> (s.begin(),s.end());
}
};