先算出每个数的pop1(twonum),pop(fivenum)然后DP ans[i][j]表示选i个数有j个2时最多有多少个5
转移方程是
for(int j=k;j>=1;j--) { for(int w=pop1;w<5000;w++) { ans[j][w]=max(ans[j][w],ans[j-1][w-pop1]+pop); } }
AC程序:
#include <bits/stdc++.h> #include <cstring> #include <iostream> #include <algorithm> #include<queue> #define EPS 1.0e-9 #define PI acos(-1.0) #define INF 30000000 #define MOD 1000000007 #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!! ") #define pb push_back #define pai pair<int,int> //using ll = long long; //using ull= unsigned long long; //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<int,vector<int>,greater<int>> que; typedef long long ll; typedef unsigned long long ull; const int maxn=1005; struct num { int two; int five; }number[205]; ll ans[205][5001]; int main() { mem(ans,-1); ans[0][0]=0; int n; int k; cin >> n >> k; int now; int pop,pop1; num temp; for(int i=1;i<=n;i++) { pop=pop1=0; cin >> now; while(now&&now%5==0) { pop++; now/=5; } while(now&&now%2==0) { pop1++; now/=2; } for(int j=k;j>=1;j--) { for(int w=pop1;w<5000;w++) { ans[j][w]=max(ans[j][w],ans[j-1][w-pop1]+pop); } } } ll anser=0; for(ll i=1;i<=5000;i++) anser=max(anser,min(i,ans[k][i])); cout<<anser<<endl; return 0; }