HDU_6033

链接

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题目描述

给一个数 $m$ ，找到一个 $k$ ，使得 $10^k\le 2^m-1$ .

思路分析

化简式子即可。
$$\begin{gathered} 10^k\le 2^m-1 \\ 10^k<2^m \\ k<lg(2^m) \\ k<m\times lg(2) \end{gathered}$$
则 $k=\lfloor m\times lg(2)\rfloor$ 。

AC代码

// #pragma GCC optimize(3)
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
// #include 
// #include 
#define endl '\n'
#define x first
#define y second
#define fi first
#define se second
#define PI acos(-1)
// #define PI 3.1415926
#define LL long long
#define INF 0x3f3f3f3f
#define lowbit(x) (-x&x)
#define PII pair<int, int>
#define ULL unsigned long long
#define PIL pair<int, long long>
#define all(x) x.begin(), x.end()
#define mem(a, b) memset(a, b, sizeof a)
#define rev(x) reverse(x.begin(), x.end())
#define IOS ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)

using namespace std;

const int N = 1e5 + 10;

int m;

void solve() {
	int T = 1;
	while (cin >> m) {
		printf("Case #%d: %d\n", T ++ , (int)floor(m * log10(2)));
	}
}

int main() {
	IOS;
	
	solve();
	
	return 0;
}
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