3.3——搜索 Many Formulas

题目描述

You are given a string S consisting of digits between 1 and 9, inclusive. You can insert the letter + into some of the positions (possibly none) between two letters in this string. Here, + must not occur consecutively after insertion.
All strings that can be obtained in this way can be evaluated as formulas.
Evaluate all possible formulas, and print the sum of the results.

Constraints
1≤|S|≤10
All letters in S are digits between 1 and 9, inclusive.

输入

The input is given from Standard Input in the following format:
S

输出

Print the sum of the evaluated value over all possible formulas.

样例输入

复制样例数据

125

样例输出

176

提示

There are 4 formulas that can be obtained: 125, 1+25, 12+5 and 1+2+5. When each formula is evaluated,

125
1+25=26
12+5=17
1+2+5=8
Thus, the sum is 125+26+17+8=176.

来源/分类

ABC045&ARC061 

ps:有点不好理解。对递归理解应该加深

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
int a[15],x;
ll ans;

void dfs(int p,ll sum){//p表示位置
    if(p==x) {
        ans+=sum;
        //cout<<"ans="<<ans<<endl;
        return;
    }
    ll t=0;
    cout<<t<<endl;
    for(int i=p;i<x;i++){//枚举每个位置，在看下一个位置，不能取+号返回。根据递归，可以得到所有情况。
        t=t*10+a[i];
        //cout<<t<<endl;
        dfs(i+1,sum+t);
    }
}
int main()
{
    ll s;
    scanf("%lld",&s);
    while(s){
        int b=s%10;
        a[x++]=b;
        s/=10;
    }
    reverse(a,a+x);
    dfs(0,0);
    cout<<ans<<endl;
    return 0;
}