1 条题解
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0
C++ :
#include<iostream> #include<string.h> #include<algorithm> using namespace std; int max(int a,long b ) { if(a>b) return a; else return b; } int main(){ int n,K,a[45]; char x; long dp[45][45]; while(cin>>n>>K){ memset(a,0,sizeof(a)); memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++){ cin>>x;a[i]=x-'0'; } for(int k=1;k<=K;k++){ for(int i=k+1;i<=n;i++){ for(int j=k;j<i;j++){ int index1=1,num1=0,index2=j+1,num2=0; while(index1<=j){ num1=num1*10+a[index1];index1++; } while(index2<=i){ num2=num2*10+a[index2];index2++; } if(k==1) dp[i][k]=max(dp[i][k],(long)num1*num2); else dp[i][k]=max(dp[i][k],dp[j][k-1]*num2); } } } cout<<dp[n][K]<<endl; } return 0; }
Pascal :
const MAXN=40; MAXK=6; var i,j,l:longint; n,K:longint; s:string; f:array[0..MAXN,0..MAXK] of int64; a:array[0..MAXN,0..MAXN] of int64; begin {assign(input,'2000t2.in');reset(input); assign(output,'2000t2.out');rewrite(output); } readln(N,K); readln(s); fillchar(f,sizeof(f),0); fillchar(a,sizeof(a),0); for i:=1 to n do begin j:=ord(s[i]); dec(j,ord('0')); a[i,i]:=j; end; for i:=1 to n do for j:=i+1 to n do a[i][j]:=a[i][j-1]*10+a[j][j]; for i:=1 to n do f[i][0]:=a[1][i]; for l:=1 to K do for i:=l+1 to n do begin for j:=l-1+1 to i-1 do if f[i][l]<f[j][l-1]*a[j+1][i] then f[i][l]:=f[j][l-1]*a[j+1][i]; end; writeln(f[N][K]); {close(input); close(output);} end.
Java :
import java.math.*; import java.util.*; public class Main { public static void main(String[] args) { Scanner cin = new Scanner ( System.in ); String str; int n ,k; while(cin.hasNext()) { n=cin.nextInt(); k=cin.nextInt(); str=cin.next(); BigInteger dp[][]=new BigInteger[41][7]; BigInteger num[][]=new BigInteger[41][41]; for(int i=1;i<=n;i++) { for(int j=i;j<=n;j++) { String s =""; for(int l=i-1;l<j;l++) { s+=str.charAt(l); } BigInteger tmp = new BigInteger(s); num[i][j] = tmp; } } BigInteger t = new BigInteger("0"); for(int i=1;i<=n;i++) { for(int j=1;j<=k;j++) { dp[i][j]=t; } } String s=""; for(int i=1;i<=n;i++) { s+=str.charAt(i-1); BigInteger tmp = new BigInteger(s); dp[i][0]=tmp; } for(int i=2;i<=n;i++) { for(int l=1;l<=k;l++) { for(int j=i-1;j>=1;j--) { dp[i][l]=dp[i][l].max(dp[j][l-1].multiply(num[j+1][i])); } } } System.out.println(dp[n][k]); } } }
- 1
信息
- ID
- 201
- 时间
- 1000ms
- 内存
- 125MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者