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Problem summary

Rearrange all digits of positive integer a to obtain the smallest positive integer b and the largest integer c. Find b+c. Leading zeros are not allowed.

Constraints

75% of tests: 0<a<10^15. The remaining tests have 10^20<a<10^50.

Worked example

An independently prepared example for this explanation.

201

312

Solution analysis

1. Read a as text, since 50 digits exceed the built-in integer types. Sort its digits increasingly for b and decreasingly for c.
2. If the increasing order starts with zero, swap its first digit with the first nonzero digit. This puts the smallest possible nonzero digit first, followed by all remaining digits in increasing order.
3. For 201 the orders become 102 and 210. Add from right to left with a carry to obtain 312. Decimal string addition also handles a final carry that increases the result length.

Why it works

The first differing digit determines numeric order among equal-length valid numbers. Therefore the greedy smallest nonzero first digit and sorted suffix minimize b, and descending digits maximize c. Standard column addition produces their exact sum.

Watch out. Do not parse the input as a 64-bit integer. All digits, including every zero, must be used exactly once.

An independently written complete program with standard headers.

#include<iostream>
#include<string>
#include<algorithm>
// Algorithm: Read a as text, since 50 digits exceed the built-in integer types.
using namespace std;
int main(){
    string smallest;
    cin>>smallest;
    sort(smallest.begin(),smallest.end());
    string largest=smallest;
    reverse(largest.begin(),largest.end());
    if(smallest[0]=='0'){
        size_t firstNonzero=smallest.find_first_not_of('0');
        swap(smallest[0],smallest[firstNonzero]);
    }
    string answer;
    int carry=0;
    for(int i=static_cast<int>(smallest.size())-1;i>=0;i--){
        int sum=smallest[i]-'0'+largest[i]-'0'+carry;
        answer.push_back(static_cast<char>('0'+sum%10));
        carry=sum/10;
    }
    if(carry!=0){
        answer.push_back(static_cast<char>('0'+carry));
    }
    reverse(answer.begin(),answer.end());
    cout<<answer<<'\n';
}