Let's call the roundness of the number the number of zeros to which it ends.
You have an array of n numbers. You need to choose a subset of exactly k numbers so that the roundness of the product of the selected numbers will be maximum possible.
The first line contains two integer numbers n and k (1 ≤ n ≤ 200, 1 ≤ k ≤ n).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 1018).
Print maximal roundness of product of the chosen subset of length k.
3 2
50 4 20
3
5 3
15 16 3 25 9
3
3 3
9 77 13
0
In the first example there are 3 subsets of 2 numbers. [50, 4] has product 200 with roundness 2, [4, 20] — product 80, roundness 1, [50, 20] — product 1000, roundness 3.
In the second example subset [15, 16, 25] has product 6000, roundness 3.
In the third example all subsets has product with roundness 0.
题目大意 给定一个数组，从中选出k个数(不能选同一个数)，使得这k个数的乘积的末尾的零的个数最多。
根据小学的数学知识，我们知道一个数的末尾有多个零取决于它质因数分解后2的指数和5的指数的最小值。(10 = 2 × 5)
所以我们初步得到动态规划的状态f[i][j][k]表示，从前i个数中，选出j个数，使得它们的乘积质因数分解后2的指数为k，5的指数最大为多少。
显然它有两种转移:选第i + 1个数，不选第i + 1个数。所以转移是显然的。
然后您会得到MLE，所以加上黑科技bfs版 + 滚动数组动态规划，不知道能不能卡过，但是有一种很简单的优化方法。
显然将题目中输入的一个数质因数分解后5的指数不会超过。所以我们可以把5个个数作为状态，2的个数作为状态的值，这样总状态数不会超过2003 * 25(刚刚是乘64)
所以继续黑科技优化内存和时间就过了。
Code
1 /** 2 * Codeforces 3 * Problem#837D 4 * Accepted 5 * Time: 187ms 6 * Memory: 10604k 7 */ 8 #include <bits/stdc++.h> 9 using namespace std; 10 typedef bool boolean; 11 #define smax(a, b) a = max(a, b); 12 template<typename T> 13 inline boolean readInteger(T& u){ 14 char x; 15 int aFlag = 1; 16 while(!isdigit((x = getchar())) && x != '-' && x != -1); 17 if(x == -1) { 18 ungetc(x, stdin); 19 return false; 20 } 21 if(x == '-'){ 22 x = getchar(); 23 aFlag = -1; 24 } 25 for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0'); 26 ungetc(x, stdin); 27 u *= aFlag; 28 return true; 29 } 30 31 typedef class Status { 32 public: 33 int stage; 34 int seced; 35 int c5; 36 }Status; 37 38 int n, k; 39 int *c2s, *c5s; 40 int res = 0; 41 42 inline void init() { 43 long long x; 44 readInteger(n); 45 readInteger(k); 46 c2s = new int[(n + 1)]; 47 c5s = new int[(n + 1)]; 48 for(int i = 1; i <= n; i++) { 49 c2s[i] = c5s[i] = 0; 50 readInteger(x); 51 while(!(x & 1)) x >>= 1, c2s[i]++; 52 while(!(x % 5)) x /= 5, c5s[i]++; 53 } 54 } 55 56 queue<Status> que; 57 int f[2][201][4001]; 58 inline void dp() { 59 int last = 0; 60 Status sta = (Status) {0, 0, 0}; 61 que.push(sta); 62 memset(f, -1, sizeof(f)); 63 f[0][0][0] = 0; 64 while(!que.empty()) { 65 Status e = que.front(); 66 que.pop(); 67 68 int lastf = f[e.stage & 1][e.seced][e.c5]; 69 Status eu = e; 70 eu.stage++; 71 if(eu.stage != last) { 72 last = eu.stage; 73 memset(f[eu.stage & 1], -1, sizeof(f[0])); 74 } 75 76 if(f[eu.stage & 1][eu.seced][eu.c5] == -1 && eu.stage < n && eu.seced <= k) 77 que.push(eu); 78 else if(eu.stage == n && eu.seced == k) 79 smax(res, min(eu.c5, lastf)); 80 smax(f[eu.stage & 1][eu.seced][eu.c5], lastf); 81 82 eu.seced++, eu.c5 += c5s[eu.stage]; 83 if(f[eu.stage & 1][eu.seced][eu.c5] == -1 && eu.stage < n && eu.seced <= k) 84 que.push(eu); 85 else if(eu.stage == n && eu.seced == k) 86 smax(res, min(eu.c5, lastf + c2s[eu.stage])); 87 smax(f[eu.stage & 1][eu.seced][eu.c5], lastf + c2s[eu.stage]); 88 } 89 } 90 91 inline void solve() { 92 printf("%d ", res); 93 } 94 95 int main() { 96 init(); 97 dp(); 98 solve(); 99 return 0; 100 }