A very hard Aoshu problem
A very hard Aoshu proble
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.
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
char str[];
int get(int i,int j,int x,int cnt)
{
int ret,a,b,c;
int s=i;
ret=a=;
for(a=;a<cnt;a++)
{
if((<<a)&x)
{
c=;
for(b=s;b<=i+a;b++)
{
c=c*+str[b]-'';
}
s=i+a+;
ret+=c; }
}
c=;
for(b=s;b<=j;b++)
c=c*+str[b]-'';
ret+=c;
return ret;
}
int main()
{
int i,j,k,ans,a,b,c,len;
while(scanf("%s",str)&&str[]!='E')
{
len=strlen(str);
ans=;
for(i=;i<len;i++)
{
for(j=;j<(<<(i-));j++)
{ for(k=;k<(<<(len-i-));k++)
{
a=get(,i-,j,i-);
b=get(i,len-,k,len-i-);
if(a==b)
{
++ans;
}
} }
}
cout<<ans<<endl;
}
return ;
}
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