たくさんの数式 / Many Formulas AtCoder

Problem Statement

 

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.

Input

 

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

S

Output

 

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

Sample Input 1

 

125

Sample Output 1

 

176

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.

Sample Input 2

 

9999999999

Sample Output 2

 

12656242944


思路：
每两个数之间的是否加“+“只有两种状态，即加和不加，那么字符串的最大长度是10，我们来用二进制思想来枚举所有状态，时间复杂度也就是2^n
显然不会TLE，用字符串substr函数来分割字符串，stringstream来转成int数字，写起来会很方便。
细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d
",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
ll getx(string str)
{
    if(str=="")
    {
        return 0ll;
    }
    stringstream ss;
    ss.clear();
    ss<<str;
    ll res;
    ss>>res;
    return res;
}
int main()
{
    //freopen("D:\common_text\code_stream\in.txt","r",stdin);
    //freopen("D:\common_text\code_stream\out.txt","w",stdout);
    string s;
    cin>>s;
    int len=s.length();
    ll ans=0ll;
    std::vector<string> v;
    for(int i=1;i<=(1<<(len-1));i++)
    {
        v.clear();
        int num=1;
        for(int j=0;j<=(len-1);j++)
        {
            if((i&(1<<j)))
            {
                string temp=s.substr(len-1-j,num);
                v.push_back(temp);
                num=1;
            }else
            {
                num++;
            }
        }
        string temp=s.substr(0,num-1);
                v.push_back(temp);
        // ll tt=0ll;
        for(auto x:v)
        {
//            db(x);
            ans+=getx(x);
        }

    }
    cout<<ans<<endl;

    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '
');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}