Codeforces Round #491 (Div. 2) F. Concise and clear

F. Concise and clear

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya is a regular participant at programming contests and is already experienced in finding important sentences in long statements. Of course, numbers constraints are important — factorization of a number less than 1000000 is easier than of a number less than 1000000000. However, sometimes it's hard to understand the number at the first glance. Could it be shortened? For example, instead of 1000000 you could write ${\mathrm{106106,\; instead\; of 1000000000 —109109,\; instead\; of 1000000007 — 109+7109+7.}}^{}$

Vasya decided that, to be concise, the notation should follow several rules:

• the notation should only consist of numbers, operations of addition ("+"), multiplication ("*") and exponentiation ("^"), in particular, the use of braces is forbidden;
• the use of several exponentiation operations in a row is forbidden, for example, writing "2^3^4" is unacceptable;
• the value of the resulting expression equals to the initial number;
• the notation should consist of the minimal amount of symbols.

Given $\mathrm{nn,\; find\; the\; equivalent\; concise\; notation\; for\; it.}$

Input

The only line contains a single integer $\mathrm{nn (1\le n\le 100000000001\le n\le 10000000000).}$

Output

Output a concise notation of the number $\mathrm{nn.\; If\; there\; are\; several\; concise\; notations,\; output\; any\; of\; them.}$

Examples

input

Copy

2018

output

Copy

2018

input

Copy

1000000007

output

Copy

10^9+7

input

Copy

10000000000

output

Copy

100^5

input

Copy

2000000000

output

Copy

2*10^9

Note

The third sample allows the answer 10^10 also of the length $55.$

 思路：10^10特判一下，之后剩的位数不超过10位，这样通过改字符来缩减长度最多出现4个字符，又单独的乘法和加法不会缩减长度，所以一定有^，于是我们将n表示成a^b*c+d的形式，a从2枚举到sqrt(n)，b从2枚举到loga(n)，贪心找最大的c，然后唯一确定一个d，这样就得到了一个sqrt(n)*log(n)复杂度的假算法，会WA9。此算法假在c或d也可能表示成e^f的形式，于是我们通过一个map预处理出所有能表示成这个形式的数，查询的时候和map里的比一下取长度最短就行了。预处理和枚举复杂度都为O(sqrt(n)*log2(n))。

  1 #include <iostream>
  2 #include <fstream>
  3 #include <sstream>
  4 #include <cstdlib>
  5 #include <cstdio>
  6 #include <cmath>
  7 #include <string>
  8 #include <cstring>
  9 #include <algorithm>
 10 #include <queue>
 11 #include <stack>
 12 #include <vector>
 13 #include <set>
 14 #include <map>
 15 #include <list>
 16 #include <iomanip>
 17 #include <cctype>
 18 #include <cassert>
 19 #include <bitset>
 20 #include <ctime>
 21 
 22 using namespace std;
 23 
 24 #define pau system("pause")
 25 #define ll long long
 26 #define pii pair<int, int>
 27 #define pb push_back
 28 #define mp make_pair
 29 #define clr(a, x) memset(a, x, sizeof(a))
 30 
 31 const double pi = acos(-1.0);
 32 const int INF = 0x3f3f3f3f;
 33 const int MOD = 1e9 + 7;
 34 const double EPS = 1e-9;
 35 
 36 /*
 37 #include <ext/pb_ds/assoc_container.hpp>
 38 #include <ext/pb_ds/tree_policy.hpp>
 39 
 40 using namespace __gnu_pbds;
 41 tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update> T;
 42 */
 43 
 44 ll n;
 45 string ans;
 46 void add(string &S, ll t) {
 47     char s[29];
 48     int l = 0;
 49     while (t) {
 50         s[++l] = t % 10;
 51         t /= 10;
 52     }
 53     while (l) {
 54         S += s[l] + '0';
 55         --l;
 56     }
 57 }
 58 map<ll, string> mmp;
 59 void pre() {
 60     for (ll a = 2; a <= sqrt(n + 0.5); ++a) {
 61         ll tt = a;
 62         for (int b = 2; ; ++b) {
 63             tt *= a;
 64             if (tt > n) break;
 65             string t2 = "";
 66             add(t2, a);
 67             t2 += "^";
 68             add(t2, b);
 69             if (mmp.count(tt)) {
 70                 string t1 = mmp[tt];
 71                 if (t2.length() < t1.length()) {
 72                     mmp[tt] = t2;
 73                 }
 74             } else {
 75                 mmp[tt] = t2;
 76             }
 77         }
 78     }
 79 }
 80 int main() {
 81     scanf("%lld", &n);
 82     if (n < 1000) {
 83         printf("%lld\n", n);
 84     } else {
 85         pre();
 86         add(ans, n);
 87         for (ll a = 2; a <= sqrt(n + 0.5); ++a) {
 88             ll tt = a;
 89             for (ll b = 2; ; ++b) {
 90                 tt *= a;
 91                 if (tt > n) break;
 92                 string res = "";
 93                 add(res, a);
 94                 res += "^";
 95                 add(res, b);
 96                 ll c = n / tt;
 97                 for (ll tc = c; tc >= max(1ll, c - 0); --tc) {
 98                     ll d = n - tt * tc;
 99                     string res2 = res;
100                     if (1 != tc) {
101                         res2 += "*";
102                         string res4 = "";
103                         add(res4, tc);
104                         if (mmp.count(tc)) {
105                             if (res4.length() < mmp[tc].length()) {
106                                 res2 += res4;
107                             } else {
108                                 res2 += mmp[tc];
109                             }
110                         } else {
111                             res2 += res4;
112                         }
113                     }
114                     if (d) {
115                         res2 += "+";
116                         string res3 = "";
117                         add(res3, d);
118                         if (mmp.count(d)) {
119                             if (res3.length() < mmp[d].length()) {
120                                 res2 += res3;
121                             } else {
122                                 res2 += mmp[d];
123                             }
124                         } else {
125                             res2 += res3;
126                         }
127                     }
128                     if (res2.length() < ans.length()) {
129                         ans = res2;
130                     }
131                 }
132             }
133         }
134         cout << ans << endl;
135     }
136     return 0;
137 }

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