A very hard Aoshu problem

最新推荐文章于 2017-11-19 00:03:09 发布

阿龙的圈子

最新推荐文章于 2017-11-19 00:03:09 发布

阅读量571

点赞数 1

CC 4.0 BY-SA版权

分类专栏：推理文章标签： ACM

本文链接：https://blog.youkuaiyun.com/DoJintian/article/details/49387133

推理专栏收录该内容

34 篇文章

订阅专栏

本文介绍了一道复杂的奥数题目，该题目要求在给定的一串数字中插入‘+’号和一个‘=’号来形成尽可能多的有效等式。通过使用递归深度优先搜索的方法遍历所有可能的等式组合，并利用哈希映射记录等式的左侧和右侧表达式的值，从而高效地解决了这个问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

D - A very hard Aoshu problem

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4403

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 <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <map>

using namespace std;
#define N 20

char s[N];
int a[N], len;
map<int, int>ml, mr;

void dfs(int pos, int x, int v, int res, map<int, int> &m)
{
    if(pos == x)
    {
        int val = res + v * 10 + a[pos];
        m[val]++;
        return ;
    }
    dfs(pos + 1, x, 0, res + v * 10 + a[pos], m);
    dfs(pos + 1, x, v * 10 + a[pos], res, m);
}

int solve(int x)
{
    int res = 0;
    ml.clear();
    mr.clear();
    dfs(0, x, 0, 0, ml);
    dfs(x + 1, len - 1, 0, 0, mr);
    for(map<int, int>::iterator it = ml.begin(); it != ml.end(); it++)
        res += (it->second) * mr[(it->first)];

    return res;
}

int main()
{
    while(~scanf("%s", s) && s[0] != 'E')
    {
        int ans = 0;
        len = strlen(s);
        for(int i = 0; i < len; i++) a[i] = s[i] - '0';
        for(int i = 0; i < len - 1; i++)
            ans += solve(i);
        printf("%d\n", ans);
    }
    return 0;
}