[Hdu 1061] Rightmost Digit (water power problems quickly divide and conquer)

Topic analysis:

Although only requires pow (x, x)% 10, but quick thinking and seeking power pow (x, y)% mod same.

The former corresponds to a simplified version of the latter.

Problem-solving ideas

1. If y is the number 1, the pow (x, 1)% mod = x% mod;
2. If an odd number of times y is 2n + 1, can be obtained pow (x, n), pow (x, y) = pow (x, n) * pow (x, n) * x
3. If y is an even number 2n, can be determined pow (x, n), pow (x, y) = pow (x, n) * pow (x, n)

The following code (G ++):

#include <bits\stdc++.h>

using namespace std;
typedef long long ll;
ll mod = 10;

//快速幂求pow(x,y)%mod
ll pow_mod(ll x, ll y) {
    //若次数y为1，则pow(x,1)%mod = x%mod;
    if (y == 1) return x;
    
    //若次数y为奇数2n+1，可求出pow(x,n)，pow(x,y) = pow(x,n)*pow(x,n)*x
    //若次数y为偶数2n，可求出pow(x,n)，pow(x,y) = pow(x,n)*pow(x,n)
    ll p = pow_mod(x, y / 2) % mod;
    if (y % 2) return (p * p * x) % mod;
    else return (p * p) % mod;
}

int main() {
    int t;
    cin >> t;
    while (t--) {
        ll n;
        cin >> n;
        cout << pow_mod(n, n) << endl;
    }
    return 0;
}