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Time Limit: 1000MS | Memory Limit: 65536K | |
Description
Input
Output
Sample Input
6 10 1 50 50 20 5
Sample Output
3650
Source
1 #include<set> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 #include<iostream> 6 #include<algorithm> 7 using namespace std; 8 const int N = 110; 9 #define For(i,n) for(int i=1;i<=n;i++) 10 #define Rep(i,l,r) for(int i=l;i<=r;i++) 11 #define Down(i,r,l) for(int i=r;i>=l;i--) 12 int n,A[N]; 13 long long dp[N][N]; 14 15 int main(){ 16 scanf("%d",&n); 17 For(i,n) scanf("%d",&A[i]); 18 For(i,n-2) dp[i][i+2]=A[i]*A[i+2]*A[i+1]; 19 Down(i,n-2,1) 20 Rep(j,i+2,n) 21 Rep(k,i+1,j-1) 22 if(dp[i][j]) 23 dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+A[i]*A[k]*A[j]); 24 else dp[i][j]=dp[i][k]+dp[k][j]+A[i]*A[k]*A[j]; 25 cout<<dp[1][n]<<endl; 26 return 0; 27 }
POJ 1651 Multiplication Puzzle
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原文地址:http://www.cnblogs.com/kingnight/p/4053785.html