HDU 4403 A very hard Aoshu problem第37回ACM/ICCC金華試合区ネット試合1004題

A very hard Aoshu problem
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 137    Accepted Submission(s): 100
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
1212 12345666 1235 END
 
 
Sample Output
2 2 0
 
 
Source
2012 ACM/ICPC Asia Regional Jinhua Online
 
 
Recommend
zhoujiaqi2010
 
 
この問題は超簡単な水問題です.
暴捜すればいい.
1つの等号、複数のプラス記号しかありません.
簡単です.
ps:今日はネットがめまいがします...明日は力を求めよう!

//1004

#include<stdio.h>

#include<iostream>

#include<map>

#include<set>

#include<algorithm>

#include<string.h>

#include<stdlib.h>

using namespace std;



int a[20];

int n;



bool solve(int s1,int s2,int t)

{

    int aa,b;

    aa=0;

    b=0;

    int temp=a[1];

    int p=1;

    for(int i=1;i<t;i++)

    {

        if(s1&(1<<(i-1)))

        {

            aa+=temp;

            temp=a[i+1];

        }

        else

        {

            temp*=10;

            temp+=a[i+1];

        }

    }

    aa+=temp;

    temp=a[t+1];

    for(int i=t+1;i<n;i++)

    {



        int qq=i-t-1;

        if(s2&(1<<qq))

        {

            b+=temp;

            temp=a[i+1];

        }

        else

        {

            temp*=10;

            temp+=a[i+1];

        }

    }

    b+=temp;

    if(aa==b)return true;

    else return false;

}

char str[30];

int main()

{

    //freopen("D.in","r",stdin);

   // freopen("D.out","w",stdout);

    while(scanf("%s",&str))

    {

        if(strcmp(str,"END")==0)break;

        int ans=0;

        n=strlen(str);

        for(int i=0;i<n;i++)

           a[i+1]=str[i]-'0';

        for(int i=1;i<n;i++)

        {

            int t1=(1<<(i-1));

            int t2=(1<<(n-i-1));

            for(int s1=0;s1<t1;s1++)

              for(int s2=0;s2<t2;s2++)

                if(solve(s1,s2,i))

                   ans++;

        }

        printf("%d
",ans);

    }

    return 0;

}