Multiplication Puzzle--动态规划

状态转移方程

dp[j][j+len]=min(dp[j][j+len],dp[j][k]+dp[k][j+len]+a[j]*a[k]*a[j+len])

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#include<stdio.h>
#include<string.h>
#define min(a,b) a>b?b:a


int a[105];
int dp[105][105];
int main()
{
int i,n;
int j,k;
while(scanf("%d",&n)!=EOF)
{
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
memset(dp,0,sizeof(dp));
for(i=0;i<n-2;i++)
dp[i][i+2]=a[i]*a[i+1]*a[i+2];
int len;
for(len=3;len<=n-1;len++)
for(j=0;j+len<n;j++)
for(k=j+1;k<j+len;k++)
{
if(dp[j][j+len]==0)
dp[j][j+len]=dp[j][k]+dp[k][j+len]+a[j]*a[k]*a[j+len];
else
dp[j][j+len]=min(dp[j][j+len],dp[j][k]+dp[k][j+len]+a[j]*a[k]*a[j+len]);
}
printf("%d\n",dp[0][n-1]);
}
return 0;
}