[BZOJ-1192]鬼谷子的钱袋

Description

读入一个数n，你要求你至少把n拆分成几个数是的随意组合后能表示1～n的任意一个数。

Solution

这是一道[HNOI2006]的题。
找到第一个大于n的2^k数，输出k。

Notice

注意是大于n的2^k数。

Code

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define sqz main
#define ll long long
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
#define per(i, a, b) for (int i = (a); i >= (b); i--)
#define Rep(i, a, b) for (int i = (a); i < (b); i++)
#define travel(i, u) for (int i = head[u]; ~i; i = edge[i].next)

const ll INF = 1e9, Mo = 998244353;
const int N = 1000000000;
const double eps = 1e-6;
namespace slow_IO
{
    ll read()
    {
        ll x = 0; int zf = 1; char ch = getchar();
        while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
        if (ch == '-') zf = -1, ch = getchar();
        while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
        return x * zf;
    }
    void write(ll y)
    {
        if (y < 0) putchar('-'), y = -y;
        if (y > 9) write(y / 10);
        putchar(y % 10 + '0');
    }
}
using namespace slow_IO;

int sqz()
{
    int n = read();
    int now = 1, cnt = 0;
    while (n > now) n -= now, cnt++, now *= 2;
    printf("%d\n", cnt + 1);
}