Table of contents

• Problem
• Time complexity
• Solution

# Problem

https://atcoder.jp/contests/abc079/tasks/abc079_c

Given 4 integers $0 \geq A, B, C, D \geq 9$, put 3 add or sub operators.
Report a combination of them whose result is 7.

# Time complexity

$O(1)$

# Solution

// C++ 14
#include <iostream>
#include <string>
#include <vector>
#include <list>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <unordered_map>
#include <math.h>

#define ll long long
#define Int int
#define loop(x, start, end) for(Int x = start; x < end; x++)
#define loopdown(x, start, end) for(int x = start; x > end; x--)
#define rep(n) for(int x = 0; x < n; x++)
#define span(a,x,y) a.begin()+x,a.begin()+y
#define span_all(a) a.begin(),a.end()
#define len(x) (x.size())
#define last(x) (*(x.end()-1))

using namespace std;

string S;
char ops[] = { '-', '+' };

void input() {
  cin >> S;
}

#define ctoi(x) (x - '0')

void solve() {
  Int x = 0;
  loop(first, 0, 2) {
    loop(second, 0, 2) {
      loop(third, 0, 2) {
        x = ctoi(S[0]);
        if (first) x += ctoi(S[1]);
        else x -= ctoi(S[1]);
        if (second) x += ctoi(S[2]);
        else x -= ctoi(S[2]);
        if (third) x += ctoi(S[3]);
        else x -= ctoi(S[3]);
        if (x == 7) {
          cout << S[0] << ops[first] << S[1] << ops[second] << S[2] << ops[third] << S[3] << "=7" << endl;
          return;
        }
      }
    }
  }
  cout << -1 << endl;
}

int main(void) {
  input();
  solve();
  return 0;
}