970.Powerful Integers

970.Powerful Integers
难度:Easy
Given two positive integers x and y, an integer is powerful if it is equal to x^i + y^j for some integers i >= 0 and j >= 0.
Return a list of all powerful integers that have value less than or equal to bound.
You may return the answer in any order. In your answer, each value should occur at most once.
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Example 1:
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Input: x = 2, y = 3, bound = 10
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Output: [2,3,4,5,7,9,10]
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Explanation: 
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2 = 2^0 + 3^0
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3 = 2^1 + 3^0
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4 = 2^0 + 3^1
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5 = 2^1 + 3^1
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7 = 2^2 + 3^1
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9 = 2^3 + 3^0
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10 = 2^0 + 3^2
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Example 2:
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Input: x = 3, y = 5, bound = 15
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Output: [2,4,6,8,10,14]
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Note:
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​
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1 <= x <= 100
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1 <= y <= 100
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0 <= bound <= 10^6
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直观暴力解法，需要考虑特殊情况，如x、y为1的情况。
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class Solution {
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vector<int> in={2};
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bool isValid(int x,int y,int num)
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{
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    if(x==1 && y==1) return num==2;
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    if(num==x+1 || num == y+1) return true;
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    int lenx=log(double(num))/log(double(x));
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  //  cout <<lenx<<endl;
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    for(int i=0;i<=lenx;i++)
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    {
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        int tmp=num-pow(x,i);
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        if(tmp==0) return false;
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       if( pow(y,int(log(double(tmp))/log(double(y)))) ==tmp)  return true;
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    }
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    return false;
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}
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public:
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    vector<int> powerfulIntegers(int x, int y, int bound) {
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        vector<int>res;
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        if(bound<2) return res;
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        if(x==1&& y==1) 
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        {
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           res.push_back(2);
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            return res;
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        }
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        if(x<y) swap(x,y);
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        if(y==1) 
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        {
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            bound-=1;
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            int lenx= log(double(bound))/log(double(x)) ;
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            for(int i=0;i<=lenx;i++)
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                res.push_back(pow(x,i)+1);
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            return res;
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        }
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        res.push_back(2);
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        for(int start=y+1;start<=bound;start++)
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            if(isValid(x,y,start))
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                res.push_back(start);
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       // cout<<isValid(x,y,bound)<<endl;
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        return res;
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    }
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};
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执行用时 :1084 ms, 在所有 C++ 提交中击败了5.60%的用户
内存消耗 :8.6 MB, 在所有 C++ 提交中击败了80.70%的用户
两重遍历，长度是对数。另外需要使用集合去重。
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class Solution {
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public:
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    vector<int> powerfulIntegers(int x, int y, int bound) {
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        if(bound<2) return vector<int>();
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        int xm= x==1 ? 1: log(double(bound))/log(double(x));
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        int ym = y==1? 1: log(double(bound))/log(double(y));
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       // cout<<xm<<" "<<ym<<endl;
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        unordered_set<int>all;
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        for(int i=0;i<=xm;i++)
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        {
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        for(int j=0;j<=ym;j++)
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        {
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            int tmp=pow(x,i)+pow(y,j);
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      //     cout<<tmp<<endl;         
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            if(tmp>bound) 
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                break;     
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            else
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                if(!all.count(tmp))all.insert(tmp);
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        }            
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        }
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        return vector<int>(all.begin(),all.end());
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    }
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};
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执行用时 :4 ms, 在所有 C++ 提交中击败了91.60%的用户
内存消耗 :8.8 MB, 在所有 C++ 提交中击败了78.07%的用户

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Last modified 5mo ago

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