平衡数
考虑将数列逆序，得到的数列使用动态规划进行计算矩阵各个元素的值。使用dp[i][j]代表从第i个位置的数字到第j个位置数字的乘积，然后对两个数列分别进行从前向后以及从后向前的对比，若存在相等的比较，则可以判断为平衡数。 
def dp():
    num = sys.stdin.readline().strip()
    
    if len(num)==0:
        print 
        print "YES"
        return
    #prod = 1
    dp = [[0 for i in range(len(num))] for j in  range(len(num))]
    rdp = deepcopy(dp)
    
    rdp = deepcopy(dp)
    for i in range(len(num)):
        dp[i][i] = int(
        dp[i][i] = int(num[i])
        rdp[i][i] = int(
        rdp[i][i] = int(num[len(num)-1-i])
        
        #prod = prod*int(num[i])
    rnum = num[::-1]
    
    for k in range(len(num)):
        
        for j in xrange(k+1,len(num)):
            dp[k][j] = dp[k][j-
            dp[k][j] = dp[k][j-1]*int(num[j])
    
    for k in range(len(num)):
        
        for j in xrange(k+1,len(num)):
            rdp[k][j] = rdp[k][j-
            rdp[k][j] = rdp[k][j-1]*int(rnum[j])
            
            #if dp[k][j]*dp[k][j]==prod:   
    for j in xrange(1,len(num)):
        
        if dp[0][j]==rdp[0][len(num)-2-j] and len(num)-2-j>=0:
                print 
                print "YES"
                return
        if j==len(num)-1:
            print 
            print "NO"