ural 1222. Chernobyl’ Eagles

1222. Chernobyl’ Eagles

Time limit: 1.0 second
Memory limit: 64 MB
Problem illustration
A Chernobyl’ eagle has several heads (for example, the eagle on the Russian National Emblem is a very typical one, having two heads; there exist Chernobyl’ eagles having twenty-six, one and even zero heads). As all eagles, Chernobyl’ eagles are very intelligent. Moreover, IQ of a Chernobyl’ eagle is exactly equal to the number of its heads. These eagles can also enormously enlarge their IQ, when they form a group for a brainstorm. IQ of a group of Chernobyl’ eagles equals to the product of IQ’s of eagles in the group. So for example, the IQ of a group, consisting of two 4-headed eagles and one 7-headed is 4*4*7=112. The question is, how large can be an IQ of a group of eagles with a given total amount of heads.

Input

There is one positive integer N in the input, N ≤ 3000 — the total number of heads of Chernobyl’ eagles in a group.

Output

Your program should output a single number — a maximal IQ, which could have a group of Chernobyl’ eagles, with the total amount of heads equal to N.

Sample

inputoutput
5
 
6
Problem Author: folklore, proposed by Leonid Volkov
Problem Source: The Seventh Ural State University collegiate programming contest
Difficulty: 141
 
题意：和为n，拆成若干个数，使得乘积最大
分析：显然，先拆成尽量多的3，在拆成2，这样乘积最大，因为肯定不能拆成1，然后之后所有数都可以用2，3构成，根据归纳法可以证明。
注意，4要拆成两个2
 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cstdlib>
 4 #include <cmath>
 5 #include <deque>
 6 #include <vector>
 7 #include <queue>
 8 #include <iostream>
 9 #include <algorithm>
10 #include <map>
11 #include <set>
12 #include <ctime>
13 using namespace std;
14 typedef long long LL;
15 typedef double DB;
16 #define For(i, s, t) for(int i = (s); i <= (t); i++)
17 #define Ford(i, s, t) for(int i = (s); i >= (t); i--)
18 #define Rep(i, t) for(int i = (0); i < (t); i++)
19 #define Repn(i, t) for(int i = ((t)-1); i >= (0); i--)
20 #define rep(i, x, t) for(int i = (x); i < (t); i++)
21 #define MIT (2147483647)
22 #define INF (1000000001)
23 #define MLL (1000000000000000001LL)
24 #define sz(x) ((int) (x).size())
25 #define clr(x, y) memset(x, y, sizeof(x))
26 #define puf push_front
27 #define pub push_back
28 #define pof pop_front
29 #define pob pop_back
30 #define ft first
31 #define sd second
32 #define mk make_pair
33 inline void SetIO(string Name) {
34     string Input = Name+".in",
35     Output = Name+".out";
36     freopen(Input.c_str(), "r", stdin),
37     freopen(Output.c_str(), "w", stdout);
38 }
39 
40 inline int Getint() {
41     int Ret = 0;
42     char Ch = ' ';
43     while(!(Ch >= '0' && Ch <= '9')) Ch = getchar();
44     while(Ch >= '0' && Ch <= '9') {
45         Ret = Ret*10+Ch-'0';
46         Ch = getchar();
47     }
48     return Ret;
49 }
50 
51 const int N = 1010, Mod = 10000;
52 int Arr[N], Len;
53 int n;
54 
55 inline void Input() {
56     scanf("%d", &n);
57 }
58 
59 inline void Mul(int x) {
60     For(i, 1, Len) Arr[i] *= x;
61     For(i, 1, Len) {
62         Arr[i+1] += Arr[i]/Mod;
63         Arr[i] %= Mod;
64     }
65     while(Arr[Len+1]) {
66         Len++;
67         Arr[Len+1] += Arr[Len]/Mod;
68         Arr[Len] %= Mod;
69     }
70 }
71 
72 inline void Solve() {
73     Arr[1] = 1, Len = 1;
74     while(n > 4) Mul(3), n -= 3;
75     if(n == 4) Mul(2), n -= 2;
76     if(n == 3) Mul(3), n -= 3;
77     if(n == 2) Mul(2), n -= 2;
78     
79     printf("%d", Arr[Len]);
80     Ford(i, Len-1, 1) printf("%04d", Arr[i]);
81     puts("");
82 }
83 
84 int main() {
85     #ifndef ONLINE_JUDGE 
86     SetIO("F");
87     #endif 
88     Input();
89     Solve();
90     return 0;
91 }
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posted @ 2015-10-19 21:25  yanzx6  阅读(214)  评论(0)  编辑  收藏  举报