P1010 a power of (divide and conquer)

https://www.luogu.com.cn/problem/P1010

 

When just see this problem, a little ignorant, if this is a math problem

7 for example, you should first obtain 7 = 4 + 2 + 1;

I.e., there should be a first exploded largest number of 2, and then down recursive

1. N is calculated how many times the power of 2 closest to the given;
2. Minus 2 to the power of n with the number of the original, if this number is greater than 2, continue to search for new n;
3. If the power is greater than 2, to perform the search power;
4. Once the number of inputs to the function 0 (exit) or 1 (0-th power of 2) or 2 (a power of two, which does not require a time out), an output;

#include <bits/stdc++.h>
using namespace std;
int n;
void search(int x){
    if(!x)
        return;
    int p = 1,q = 0;
    cout << 2;
    while(x >= p){
        p *= 2;
        q++;
    }
    q--;
    if(!q || q == 2)
        cout << "(" << q << ")";
    if(q >= 3){
        cout << "(";
        search(q);
        cout << ")";
    }
    x -= p / 2;
    if(x){
        cout << "+";
        search(x);
    }

}
int main(){
    ios::sync_with_stdio(0);
    cin >> n;
    search(n);
    return 0;
}

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