动态规划Round Subset
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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 withroundness 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:从N个数中选出M个数使得这M个数的乘积后的0最多。
试题分析:不难发现,构成一个0的条件是2*5,那么对于每一个数字我们求出它的质因数分解中有多少2多少5
dp[i][j]表示选i个数其中有j个2最多有多少个5
那么dp[i][j]=max(dp[i-1][j-t2]+t5);
其中t2为质因数分解中2的个数,t5为质因数分解中5的个数。
代码:
#include<iostream> #include<cstring> #include<cstdio> #include<vector> #include<queue> #include<cmath> #include<stack> #include<algorithm> using namespace std; inline long long read(){ long long x=0,f=1;char c=getchar(); for(;!isdigit(c);c=getchar()) if(c==‘-‘) f=-1; for(;isdigit(c);c=getchar()) x=x*10+c-‘0‘; return x*f; } const int MAXN=1001; const int INF=0x3f3f3f; const int n2=206*64; int MAX=-INF; int N,M; int dp[MAXN][n2+1]; long long a[MAXN]; int ans; int main(){ N=read(),M=read(); for(int i=1;i<=N;i++) a[i]=read(); for(int i=0;i<=M;i++) for(int j=0;j<n2;j++) dp[i][j]=-INF; dp[0][0]=0; for(int i=1;i<=N;i++){ long long x=a[i],x2=a[i]; int T2=0,T5=0; while(x%2==0) x/=2,T2++; while(x2%5==0) x2/=5,T5++; for(int k=M;k>=1;k--) for(int j=T2;j<n2;j++) dp[k][j]=max(dp[k-1][j-T2]+T5,dp[k][j]); } ans=0; for(int i=1;i<n2;i++) ans=max(ans,min(dp[M][i],i)); cout<<ans<<endl; }
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