How Many Equations Can You Find

最新推荐文章于 2025-09-02 20:53:00 发布

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本文介绍了一种通过深度优先搜索(DFS)算法解决特定字符串加减运算的问题。给定一个仅包含数字的字符串和一个整数目标值N，文章探讨了如何在数字间插入加号或减号使表达式的计算结果等于N的方法总数。通过对每一步进行选择，并回溯到上一步继续尝试不同路径的方式，最终统计并输出所有可行方案的数量。

题目来源2017HPUACM暑期培训：https://vjudge.net/contest/174968#problem/E

 Now give you an string which only contains 0, 1 ,2 ,3 ,4 ,5 ,6 ,7 ,8 ,9.You are asked to add the sign ‘+’ or ’-’ between the characters. Just like give you a string “12345”, you can work out a string “123+4-5”. Now give you an integer N, please tell me how many ways can you find to make the result of the string equal to N .You can only choose at most one sign between two adjacent characters.

Input
Each case contains a string s and a number N . You may be sure the length of the string will not exceed 12 and the absolute value of N will not exceed 999999999999.
Output
The output contains one line for each data set : the number of ways you can find to make the equation.
Sample Input

123456789 3
21 1

Sample Output

18
1

dfs的运用

#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#define ll long long
using namespace std;
ll n,ans;
int len;
char s[15];
void dfs(ll sum,ll x)//s时当前的和，x时现在的字符位置 
{
    if(x==len)
    {
        if(sum==n)
        ans++;
        return;
    }
    ll a=0;
    for(int i=x;i<len;i++)
    {
        a=a*10+s[i]-'0';
        dfs(sum+a,i+1);//接下来的几行时判断+ -号哪个符合 
        if(x!=0)
        {
            dfs(sum-a,i+1);
        }
    }
}
int main()
{
    while(~scanf("%s %lld",s,&n))
    {
        len=strlen(s);
        ans=0;
        dfs(0,0);
        printf("%lld\n",ans);
    }
return 0;
}