bzoj1192 鬼谷子的钱袋

题意：把一个数n进行拆分，拆分出来大于一的数两两不等，使得拆出来的数可以组成[1, n]间的所有数。求最少拆成多少个数。n <= 1000000000。

@Azrael_Death推荐的水题

拆出来的数越小越好，又要组成所有数，易想到拆分二进制。求n在二进制下的位数即可。

#include<map>#include<cmath>#include<cstdio>#include<queue>#include<vector>#include<string>#include<cstring>#include<iostream>#include<algorithm>using namespace std;inline int read() {int x = 0, flag = 1; char ch = getchar();while (ch > '9' || ch < '0') { if (ch == '-') flag = -1; ch = getchar(); }while (ch <= '9' && ch >= '0') { x = x * 10 + ch - '0'; ch = getchar(); }return x * flag;}#define rep(ii, aa, bb) for (int ii = aa; ii <= bb; ii++)#define drp(ii, aa, bb) for (int ii = aa; ii >= bb; ii--)#define ll long longint main() {int m; cin >> m;int cnt = 0;while (m) m >>= 1, cnt++;cout << cnt;return 0;}