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=20−9−8=20−9, 7=24−97=24−9, 1015=210−91015=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 (24−9)(24−9).
Note that negative pp-binary numbers are allowed to be in the sum (see the Notes section for an example).
The only line contains two integers nn and pp (1≤n≤1091≤n≤109, −1000≤p≤1000−1000≤p≤1000).
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.
24 0
2
24 1
3
24 -1
4
4 -7
2
1 1
-1
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=(24−1)+(22−1)+(22−1)+(22−1)24=(24−1)+(22−1)+(22−1)+(22−1). Note that repeated summands are allowed.
In the fourth sample case, we can represent 4=(24−7)+(21−7)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,若 尝试的个数 小于 等于全为1所需要的个数 且 大于单纯拆分最少的个数 则这个尝试个数有效
#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; }