#638992 (C++17) No.8077 🔧

• 問題: No.8077 🔧
• ユーザー: kanra824
• 提出日時: 2021-04-01 20:17:28
• 言語: C++17 (gcc 13.3.0 + boost 1.87.0)
• 結果: AC
• 実行時間: 2 ms / 2,000 ms
• コード長: 1,948 bytes
• コンパイル時間: 1,926 ms
• コンパイル使用メモリ: 191,992 KB
• 最終ジャッジ日時: 2025-01-20 05:27:50
• ジャッジサーバーID （参考情報）: judge5 / judge5

ファイルパターン	結果
sample	AC * 3
other	AC * 10

ソースコード

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#include <bits/stdc++.h>
using namespace std;
#define REP(i, n) for(int i=0; i<(n); ++i)
#define RREP(i, n) for(int i=(n);i>=0;--i)
#define FOR(i, a, n) for (int i=(a); i<(n); ++i)
#define RFOR(i, a, b) for(int i=(a);i>=(b);--i)
#define SZ(x) ((int)(x).size())
#define ALL(x) (x).begin(),(x).end()
#define DUMP(x) cerr<<#x<<" = "<<(x)<<endl
#define DEBUG(x) cerr<<#x<<" = "<<(x)<<" (L"<<__LINE__<<")"<<endl;
template<class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  REP(i, SZ(v)) {
    if (i) os << " ";
    os << v[i];
  }
  return os;
}
template <class T>
void debug(const vector<T> &v) {
  cout << "[";
  REP(i, SZ(v)) {
    if(i) cout << ", ";
    cout << v[i];
  }
  cout << "]" << endl;
}
template<class T, class U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  return os << p.first << " " << p.second;
}
template <class T, class U>
void debug(const pair<T, U> &p) {
  cout << "(" << p.first << " " << p.second << ")" << endl;
}
template<class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template<class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using P = pair<int, int>;
using vi = vector<int>;
using vll = vector<ll>;
using vvi = vector<vi>;
using vvll = vector<vll>;
const ll MOD = 1e9 + 7;
const ll MOD998 = 998244353;
const int INF = INT_MAX;
const ll LINF = LLONG_MAX;
const int inf = INT_MIN;
const ll linf = LLONG_MIN;
const ld eps = 1e-9;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  ll a, b, c, d, k; cin >> a >> b >> c >> d >> k;
  cout << (a * b + c - d) % k << endl;
  return 0;
}
 
 
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