2012年9月22日金华网络预赛 D题结题报告——子集划分，数学

最新推荐文章于 2017-11-19 00:03:09 发布

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#网络 #iterator #class #output #integer #input

本文探讨了Aoshu数学难题的解决方法，通过子集划分技巧求解给定数字序列中可能形成的等式数量。适用于初级数学爱好者。

A very hard Aoshu problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0

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

Sample Output

2
2
0

#include <algorithm>

#include <iostream>
#include <iomanip>
#include <string>

#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>

#include <vector>
#include <stack>
#include <queue>
#include <map>
#include <utility>
#include <set>
#include <bitset>

using namespace std;

char s[100];

template<
    class Key,
    class Compare = std::less<Key>,
    class Allocator = std::allocator<Key>
> class multiset;


map <__int64, int > s1;
map <__int64, int > s2;

__int64 mypow(int x, int y)
{
    __int64 t;
    if (y == 0)
        return 1;
    else if (y == 1)
        return x;
    else if (y == 2)
        return x * x;
    else if (y % 2)
    {
        t = mypow(x, (y - 1) / 2);
        return t * t * x;
    }

    else
    {
        t = mypow(x ,y / 2);
        return t * t;
    }
}

void f(map <__int64, int> &ss, int l, int h)
{
    int i, j, k, p, t[100];
    __int64 num, n;

    n = 0;
    p = 0;
    for (i = (1 << (h - l)) - 1; i >= 0; i--)
    {
        num = 0;
        for (j = 0; j < (h - l); j++)
        {
            if (i & (1 << j))
            {
                n = 0;
                t[p++] = (int) (s[h - j] - 48);
                for (k = 0; k < p; k++)
                {
                    n += (t[k] * mypow(10, k));
                }
                num += n;
                n = 0;
                p = 0;
            }

            else
            {
                t[p] =  (int)(s[h-j]  - 48);
                p++;
            }
        }

        t[p++] = (int)(s[l]  - 48);

        for (k = 0; k < p; k++)
        {
            n += (t[k] * mypow(10,  k));
        }
        num+= n;
//        ss.insert(num);
        ss[num]++;
        p = 0;
        n = 0;
        num = 0;
    }

}


int main()
{
    int ans, len, j;
    map <__int64, int> :: iterator it;
    map <__int64, int> :: iterator it2;
    scanf("%s", s);
    while ( strcmp(s, "END") != 0)
    {
        ans = 0;
        len = strlen(s);

        for (j = 1; j <= len - 1; j++)
        {
            s1.clear();
            s2.clear();
            f(s1, 0, j - 1);

            f(s2, j, len - 1);

            for (it = s1.begin(); it != s1.end(); it++)
            {

                for (it2 = s2.begin(); it2 != s2.end(); it2++)
                {
                    if (it ->first == it2 -> first)
                    {
                        ans += it -> second * it2 -> second;
                        break;
                    }
                }




            }
        }

        printf("%d\n", ans);
        scanf("%s", s);
    }

    //system("pause");
    return 0;
}