LA - Chop Ahoy! Revisited!

Given a non-empty string composed of digits only, we may group these digits into sub-groups (but maintaining their original order) if, for every sub-group but the last one, the sum of the digits in a sub-group is less than or equal to the sum of the digits in the sub-group immediately on its right. Needless to say, each digit will be in exactly one sub-group. For example, the string 635 can only be grouped in one sub-group [635] or in two sub-groups as follows: [6-35] (since 6 < 8.) Another example is the string 1117 which can be grouped in one sub-group [1117] or as in the following: [1-117], [1-1-17], [1-11-7], [1-1-1-7], [11-17], and [111-7] but not any more, hence the total number of possibilities is 7. Write a program that computes the total number of possibilities of such groupings for a given string of digits.

Input
Your program will be tested on a number of test cases. Each test case is specified on a separate line. Each line contains a single string no longer than 25, and is made of decimal digits only. The end of the test cases is identified by a line made of the word "bye" (without the quotes.) Such line is not part of the test cases.

Output
For each test case, write the result using the following format: k._n where k is the test case number (starting at 1,) and n is the result of this test case.

Analyses

 * View problem analysis (written by Mido1236)

/*ba yade oo */


 * 1) include
 * 2) include
 * 3) include 
 * 4) include
 * 5) include 
 * 6) include 
 * 7) include
 * 8) include
 * 9) include
 * 10) include

using namespace std;


 * 1) define PI 3.14159265358997
 * 2) define absol(x) ((x)>(0) ? (x):(-1)*(x))
 * 3) define pow2(x) (x*x)
 * 4) define EPS 1e-7
 * 5) define INF 0x7FFFFFFF
 * 6) define MAX 30

int num[MAX],sum[MAX]; int t[MAX][MAX*10]; int main {

char ch[MAX]; int len; int Case=1; while (cin>>ch) {		if (!strcmp(ch,"bye")) break; len=strlen(ch); sum[0]=0; for (int i=0;i=0;j--) {					int maxi=sum[i]-sum[j]; if (k>=maxi) t[i][k]+=t[j][maxi]; }			}

printf("%d. %d\n",Case++,t[len-1][sum[len-1]]); }	return 0;

}