C - たくさんの数式 / Many Formulas

字符串数式求和算法

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本文介绍了一种解决特定字符串数式求和问题的算法，通过二进制思想来枚举所有可能的加法组合，并计算这些组合的总和。算法使用C++实现，能够高效地处理输入字符串。

C - たくさんの数式 / Many Formulas

Time limit : 2sec / Memory limit : 256MB

Score : 300 points

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:

Output

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

Sample Input 1

Copy

Sample Output 1

Copy

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

Copy

9999999999

Sample Output 2

Copy

12656242944

题解：二进制思想。1代表有‘+‘，0代表没有。

#include <iostream>
#include <cstdio>
#include <string>
using namespace std;

long long toIntSum(string s)
{
    long long sum=0,tmp=0;
    for (int i=0;i<s.size();i++)
    {
        if(s[i]=='+')
        {
           sum+=tmp;
           tmp=0;
        }
        else
            tmp=tmp*10+s[i]-'0';

    }
    return sum+tmp;
}
int main ()
{
    string str;
    cin>>str;
    int len =str.size()-1,a[11]={0};//加号最多=串长-1;a[i]==0表示i处没有加号，a[i]==1表示i处有加号.
    long long sum=0;
    while(true)
    {
        string s;
        for (int i=0;i<len;i++)
        {
            a[i+1]+=a[i]/2;
            a[i]%=2;
        }
        if(a[len])
            break;
        for (int i=0;i<len;i++)//把加号和数字合成一个串
        {
            s+=str[i];
            if(a[i])
                s+="+";
        }
        s+=str[len];
        sum+=toIntSum(s);
        a[0]++;
    }
    cout<<sum;
    return 0;
}