372. 超级次方

快速幂超时 因为要转二进制

typedef long long LL;

class Solution {
public:
    vector<int> chu2(vector<int>& b){

        int jie = 0;
        for(int i=0;i<b.size();i++){
            int curr = jie*10 + b[i];
            jie = curr % 2;
            b[i] = curr/2;
        }
        return b;
    }
    bool iszero(vector<int>& b){
        for(auto i:b){
            if(i!=0){
                return false;
            }
        }
        return true;
    }
    string to_bin(vector<int> b){
        string bin = "";
        while(!iszero(b)){
            bin += b[b.size()-1]%2+'0';
            b = chu2(b);

        }
        for(int i=0;i < bin.size()/2;i++){
            swap(bin[i],bin[bin.size()-1-i]);
        }
        //cout<<"bin "<<bin<<endl;
        return bin;
    }
    int superPow(int a, vector<int>& b) {
        LL x = a;
        //x %= 1337;
        LL ans = 1;
        string bin = to_bin(b);
        int index = bin.size()-1;
        while(index>=0){
            if(bin[index]=='1'){
                ans = ans*x %1337;
            }
            x = (x*x)%1337;
            index--;
        }
        return ans;
    }
};

递归：
$a^{xyz} = (a^{xy})^{10}*a^{z}$axyz=(axy)10∗az
计算一位的时候用快速幂 因为过程中要取模 否则超范围。

class Solution {
public:
    int myPow(int a,int n){
        int ans = 1;
        a%=1337;
        while(n){
            if(n&1){
                ans*=a;
                ans%=1337;
            }
            a*=a;
            a%=1337;
            n>>=1;
        }
        return ans;
    }
    int superPow(int a, vector<int>& b) {
        if(a==0 || a==1){
            return a;
        }else if(b.size() == 1 ){
            return int ( myPow(a,b[0]) )%1337 ;
        }
        else{
            int ba = b.back();
            b.pop_back();
            return  ( int(myPow( superPow(a,b),10)) %1337 * int(myPow(a,ba))%1337 )%1337;
        }
    }
};

欧拉降幂
$a^b\equiv \left\{\begin{matrix} a^b\% & 2\\ & \\ & \end{matrix}\right.$ab≡⎩⎨⎧​ab%​2​
模运算可以分配

class Solution {
public:
    int myPow(int a,int n){
        int ans = 1;
        a%=1337;
        while(n){
            if(n&1){
                ans*=a;
                ans%=1337;
            }
            a*=a;
            a%=1337;
            n>>=1;
        }
        return ans;
    }

    int superPow(int a, vector<int>& b) {
        int n = 0;
        if(b.size()==0){
            return 0;
        }
        for(int i=0;i<b.size();i++){
            n*=10;
            n+=b[i];
            n%=1140;
        }
        return myPow(a,n);
    }
};

标签：bin,return,int,超级,ans,次方,372,1337,size
来源： https://blog.csdn.net/L1558198727/article/details/89371783