spoj-ANARC05H -dp

ANARC05H - Chop Ahoy! Revisited!

#dynamic-programming

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.

Example

Input:
635
1117
9876
bye

Output:
1. 2
2. 7
3. 2
      f[i][j]表示进行到第i位，且最后一个集合包含j个数字的方案个数，显然f[i][j]=SUM{f[i-j][k]  | 只要最后i的最后j个数的和>=i-j的最后k个数的和就能转移}

#include<bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
int main()
{
    char s[30];
    int x[30]={0};
    int f[30][30];
    int i,j,k,cas=0;
    while(scanf("%s",s+1)){
        if(!strcmp(s+1,"bye")) break;
        int n=strlen(s+1);
        for(i=1;i<=n;++i){
            x[i]=x[i-1]+s[i]-'0';
        }
        memset(f,0,sizeof(f));
        f[0][0]=f[1][1]=1;
        for(i=2;i<=n;++i){
            for(j=1;j<=i;++j){
                if(i==j) f[i][j]=1;
                else f[i][j]=0;
                for(k=1;k<=i-j;++k){
                    if(x[i]-x[i-j]>=x[i-j]-x[i-j-k])
                     f[i][j]+=f[i-j][k];
                }
            }
        }
        int s=0;
        for(i=0;i<=n;++i) s+=f[n][i];
        cout<<++cas<<". "<<s<<endl;
    }
    return 0;
}