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(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 ‘\n‘ #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 */
(Incomplete)UVa 701 The Archeologist's Dilemma
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