(Incomplete)UVa 701 The Archeologist's Dilemma

方法：对数，暴力

我们需要求出最小的e，是的存在一个i > len(n) , 满足   floor[2^e/ (10^i)]= n, 即 n*10^i < 2^e < (n+1)*10^i。对两边同时取log10 （以10为底的对数，记作lg），得到

lg(n) + i < e*lg(2) < lg(n+1) + i。 注意len(n) = (int) lg(n)+1, 之前一个条件 i > len(n) 可以转化为 i > lg(n)+1,  然后暴力枚举即可。 注意2的power有无限个，无解的情况不存在（待考证）。

code:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#include <fstream>
#include <cassert>
#include <unordered_map>
#include <cmath>
#include <sstream>
#include <time.h>
#include <complex>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define FOR(a,b,c) for (int (a)=(b);(a)<(c);++(a))
#define FORN(a,b,c) for (int (a)=(b);(a)<=(c);++(a))
#define DFOR(a,b,c) for (int (a)=(b);(a)>=(c);--(a))
#define FORSQ(a,b,c) for (int (a)=(b);(a)*(a)<=(c);++(a))
#define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a))
#define FOREACH(a,b) for (auto &(a) : (b))
#define rep(i,n) FOR(i,0,n)
#define repn(i,n) FORN(i,1,n)
#define drep(i,n) DFOR(i,n-1,0)
#define drepn(i,n) DFOR(i,n,1)
#define MAX(a,b) a = Max(a,b)
#define MIN(a,b) a = Min(a,b)
#define SQR(x) ((LL)(x) * (x))
#define Reset(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(v) v.begin(),v.end()
#define ALLA(arr,sz) arr,arr+sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(all(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr,sz) sort(ALLA(arr,sz))
#define REVERSEA(arr,sz) reverse(ALLA(arr,sz))
#define PERMUTE next_permutation
#define TC(t) while(t--)
#define forever for(;;)
#define PINF 1000000000000
#define newline '
'

#define test if(1)if(0)cerr
using namespace std;
  using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> ii;
typedef pair<double,double> dd;
typedef pair<char,char> cc;
typedef vector<ii> vii;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> l4;
const double pi = acos(-1.0);

int n;
int solve()
{
    int i = log10(n)+2;
    for (;;++i)
    {
        int a = (log10(n) + i)/log10(2), b = ceil((log10(n+1)+i)/log10(2));
        
        if (a+1 < b)
            return a+1;
    }
    return -1;
}

int main()
{
    while (cin >> n)
    {
        cout << solve() << newline;
    }
}
/*
 
 0 0 0 0
 0 0 0 100
 5 20 34 325 4 5 6 7
 283 102 23 320 203 301 203 40 -1 -1 -1 -1
*/