• C. p-binary


    Vasya will fancy any number as long as it is an integer power of two. Petya, on the other hand, is very conservative and only likes a single integer pp (which may be positive, negative, or zero). To combine their tastes, they invented pp-binary numbers of the form 2x+p2x+p, where xx is a non-negative integer.

    For example, some 9−9-binary ("minus nine" binary) numbers are: 8−8 (minus eight), 77 and 10151015 (8=209−8=20−9, 7=2497=24−9, 1015=21091015=210−9).

    The boys now use pp-binary numbers to represent everything. They now face a problem: given a positive integer nn, what's the smallest number of pp-binary numbers (not necessarily distinct) they need to represent nn as their sum? It may be possible that representation is impossible altogether. Help them solve this problem.

    For example, if p=0p=0 we can represent 77 as 20+21+2220+21+22.

    And if p=9p=−9 we can represent 77 as one number (249)(24−9).

    Note that negative pp-binary numbers are allowed to be in the sum (see the Notes section for an example).

    Input

    The only line contains two integers nn and pp (1n1091≤n≤109, 1000p1000−1000≤p≤1000).

    Output

    If it is impossible to represent nn as the sum of any number of pp-binary numbers, print a single integer 1−1. Otherwise, print the smallest possible number of summands.

    Examples
    input
    Copy
    24 0
    
    output
    Copy
    2
    
    input
    Copy
    24 1
    
    output
    Copy
    3
    
    input
    Copy
    24 -1
    
    output
    Copy
    4
    
    input
    Copy
    4 -7
    
    output
    Copy
    2
    
    input
    Copy
    1 1
    
    output
    Copy
    -1
    
    Note

    00-binary numbers are just regular binary powers, thus in the first sample case we can represent 24=(24+0)+(23+0)24=(24+0)+(23+0).

    In the second sample case, we can represent 24=(24+1)+(22+1)+(20+1)24=(24+1)+(22+1)+(20+1).

    In the third sample case, we can represent 24=(241)+(221)+(221)+(221)24=(24−1)+(22−1)+(22−1)+(22−1). Note that repeated summands are allowed.

    In the fourth sample case, we can represent 4=(247)+(217)4=(24−7)+(21−7). Note that the second summand is negative, which is allowed.

    In the fifth sample case, no representation is possible.

     拆分二进制, 尝试的个数 小于 等于全为1所需要的个数 大于单纯拆分最少的个数 这个尝试个数有效

    #include <iostream>
    #include <vector>
    #include <algorithm>
    #include <string>
    #include <set>
    #include <queue>
    #include <map>
    #include <sstream>
    #include <cstdio>
    #include <cstring>
    #include <numeric>
    #include <cmath>
    #include <iomanip>
    #include <deque>
    #include <bitset>
    #include <unordered_set>
    #include <unordered_map>
    #define ll              long long
    #define PII             pair<int, int>
    #define rep(i,a,b)      for(int  i=a;i<=b;i++)
    #define dec(i,a,b)      for(int  i=a;i>=b;i--)
    using namespace std;
    int dir[4][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 } };
    const long long INF = 0x7f7f7f7f7f7f7f7f;
    const int inf = 0x3f3f3f3f;
    const double pi = 3.14159265358979323846;
    const double eps = 1e-6;
    const int mod = 998244353;
    const int N = 1e6 + 5;
    //if(x<0 || x>=r || y<0 || y>=c)
    
    inline ll read()
    {
        ll x = 0; bool f = true; char c = getchar();
        while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
        while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
        return f ? x : -x;
    }
    ll gcd(ll m, ll n)
    {
        return n == 0 ? m : gcd(n, m % n);
    }
    ll lcm(ll m, ll n)
    {
        return m * n / gcd(m, n);
    }
    
    int main()
    {
        ll n, p;
        cin >> n >> p;
        for (int i = 0; i <= 1e6; i++)
        {
            int cnt=0,tmp = n - i*p;
            while (tmp)
            {
                cnt += tmp % 2;
                tmp /= 2;
            }
              if (cnt <= i && i <= n - i * p)
            {
                cout << i << endl;
                return 0;
            }
        }
        cout << -1 << endl;
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/dealer/p/13037778.html
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