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49 lines
1.1 KiB
Markdown
49 lines
1.1 KiB
Markdown
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# [372. Super Pow](https://leetcode.com/problems/super-pow/)
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# 思路
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题目让实现一个超级pow函数,最后结果对1337取模。
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先不考虑对1337取模的话实现pow函数是很简单的,就是一个二分的思想:
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``` C++
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int pow(int x, int y){ // x >=0, y >= 0
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if(y == 0) return 1;
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if(y == 1) return x;
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return pow(x, y/2) * pow(x, y - y/2);
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}
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```
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由于此题的结果可能很大,需要对1337取模,所以我们在计算的过程中能对1377取模的地方就对1337取模:
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``` C++
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int pow(int x, int y){ // (x^y) % 1337
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if(y == 0) return 1;
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x %= 1337;
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if(y == 1) return x;
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return pow(x, y/2) * pow(x, y - y / 2) % 1337;
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}
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```
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剩下就好写了,见下面完整代码。
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# C++
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``` C++
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class Solution {
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private:
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int pow(int x, int y){ // (x^y) % 1337
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if(y == 0) return 1;
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x %= 1337;
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if(y == 1) return x;
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return pow(x, y/2) * pow(x, y - y / 2) % 1337;
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}
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public:
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int superPow(int a, vector<int>& b) {
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int res = 1;
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for(int i = 0; i < b.size(); i++)
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res = pow(res, 10) * pow(a, b[i]) % 1337;
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return res;
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}
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};
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```
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