2022 Nioke winter vacation algorithm basic training camp 2, Xiaosha's counting (greedy, bit operation)

portal

Ideas:

Since there is more than one one in the case of binary splitting and the last parity, then the contribution after XOR must be less than our cost, so we have to ensure that the number of each bit is either 0 or 1, so that we can guarantee $a\left[\wedge \right]=a\left[+\right]$（因为$a+b=2\ast \left(a\&b\right)+a\wedge b\ge a\wedge b$，所以$a\left[\wedge \right]=m$ is the optimal case)

Then we found that for each of them, they do not interfere with each other, then the situation they may produce is $n$ types, so we only need to ask for mmunder binary$How\; many\; 1s\; does\; m$ have, then find$n^x$ is enough , and the time complexity is$O\left({\log }_{2}m\right)$

Code:

#include<iostream>
using namespace std;
const int mod = 1e9+7;
int main(){
    
    
    long long n, m, ans = 1;
    cin >> n >> m;
    n %= mod;
    while(m){
    
    
        if(m&1)	ans = ans * n % mod;
        m >>= 1;
    }
    cout << ans << endl;
    return 0;
}