HDU4403-A very hard Aoshu problem

A very hard Aoshu problem

                                                                        Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                                   Total Submission(s): 1520    Accepted Submission(s): 1041

Problem Description

Aoshu is very popular among primary school students. It is mathematics, but much harder than ordinary mathematics for primary school students. Teacher Liu is an Aoshu teacher. He just comes out with a problem to test his students:

Given a serial of digits, you must put a '=' and none or some '+' between these digits and make an equation. Please find out how many equations you can get. For example, if the digits serial is "1212", you can get 2 equations, they are "12=12" and "1+2=1+2". Please note that the digits only include 1 to 9, and every '+' must have a digit on its left side and right side. For example, "+12=12", and "1++1=2" are illegal. Please note that "1+11=12" and "11+1=12" are different equations.

 

Input

There are several test cases. Each test case is a digit serial in a line. The length of a serial is at least 2 and no more than 15. The input ends with a line of "END".

 

Output

For each test case , output a integer in a line, indicating the number of equations you can get.

 

Sample Input

1212123456661235END

 

Sample Output

220

 

Source

2012 ACM/ICPC Asia Regional Jinhua Online

 

题意：给定一串数字，在这些数字中插入一些‘+’和一个’=’ 使得两边相等，求有几种方式

解题思路：先预处理出每一段的数值，在数值中间放登号，然后dfs

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <stack>#include <vector>#include <set>#include <map>using namespace std;#define LL long longint ans;int x[20][20];char ch[20];int len;void init(){    memset(x,0,sizeof x);    for(int i=0;i<len;i++)    {        for(int j=i;j<len;j++)        {            for(int k=i;k<=j;k++)                x[i][j]=x[i][j]*10+(ch[k]-'0');        }    }}void dfs(int k,int pos,int sum1,int sum2){    if(pos>=len)    {        if(sum1==sum2)        {            ans++;return ;        }    }    if(pos<=k)    {        for(int i=pos;i<=k;i++)            dfs(k,i+1,sum1+x[pos][i],sum2);    }    else    {        for(int i=pos;i<len;i++)            dfs(k,i+1,sum1,sum2+x[pos][i]);    }}int main(){    while(~scanf("%s",ch)&&(ch[0]!='E'))    {        ans=0;        len=strlen(ch);        init();        for(int i=0;i<len-1;i++)            dfs(i,0,0,0);        printf("%d\n",ans);    }    return 0;}