open back contain msu () put val atom ready

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

106106, instead of 1000000000 —${\mathrm{109109,\; instead\; of\; 1000000007\; —\; 109+7109+7.}}^{}$

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

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，之後剩的位數不超過10位，這樣通過改字符來縮減長度最多出現4個字符，又單獨的乘法和加法不會縮減長度，所以一定有^，於是我們將n表示成a^b*c+d的形式，a從2枚舉到sqrt(n)，b從2枚舉到loga(n)，貪心找最大的c，然後唯一確定一個d，這樣就得到了一個sqrt(n)*log(n)復雜度的假算法，會WA9。此算法假在c或d也可能表示成e^f的形式，於是我們通過一個map預處理出所有能表示成這個形式的數，查詢的時候和map裏的比一下取長度最短就行了。預處理和枚舉復雜度都為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 }

View Code

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