Submission #1986310

---

Source Code Expand

#include <iostream>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <algorithm>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cmath>
using namespace std;

using ll = long long;
using P = pair<int, int>;

const int INF = 1e9;
const int MOD = 1e9 + 7;

// 4近傍、8近傍
int dx[] = {1, 0, -1, 0, 1, -1, -1, 1};
int dy[] = {0, 1, 0, -1, 1, 1, -1, -1};

int ans[3][3];

bool check() {
  // たて、横、斜めの和をsetに入れていき、最終的にサイズが1ならOK
  set<int> st;
  int sum;
  for (int i = 0; i < 3; i++) {
    sum = 0;
    for (int j = 0; j < 3; j++) {
      sum += ans[i][j];
    }
    st.insert(sum);
  }
  for (int i = 0; i < 3; i++) {
    sum = 0;
    for (int j = 0; j < 3; j++) {
      sum += ans[j][i];
    }
    st.insert(sum);
  }
  sum = 0;
  for (int i = 0, j = 0; i < 3; i++, j++) {
    sum += ans[i][j];
  }
  st.insert(sum);
  sum = 0;
  for (int i = 2, j = 0; i >= 0; i--, j++) {
    sum += ans[i][j];
  }
  st.insert(sum);

  return st.size() == 1;
}

int main() {
  // 入力
  int A, B, C;
  cin >> A >> B >> C;

  // 総和として0〜300の候補があるので、全探索
  ans[0][0] = A;
  ans[0][1] = B;
  ans[1][1] = C;
  for (int s = 0; s <= 300; s++) {
    ans[0][2] = s - (A + B);
    ans[2][1] = s - (B + C);
    ans[2][2] = s - (A + C);
    ans[1][2] = s - (ans[0][2] + ans[2][2]);
    ans[2][0] = s - (ans[0][2] + ans[1][1]);
    ans[1][0] = s - (ans[1][1] + ans[1][2]);

    if (check()) break;
  }

  // 解答
  for (int i = 0; i < 3; i++) {
    for (int j = 0; j < 3; j++) {
      cout << ans[i][j];
      if (j < 2) cout << " ";
    }
    cout << endl;
  }

  return 0;
}

Submission Info

Judge Result