結果

問題 No.967 引き算をして門松列(その2)
ユーザー kissshot7kissshot7
提出日時 2020-07-21 01:40:12
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 23 ms / 2,000 ms
コード長 3,833 bytes
コンパイル時間 1,736 ms
コンパイル使用メモリ 170,752 KB
実行使用メモリ 7,748 KB
最終ジャッジ日時 2024-06-07 09:39:07
合計ジャッジ時間 2,438 ms
ジャッジサーバーID
(参考情報)
judge4 / judge1
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 4 ms
7,620 KB
testcase_01 AC 4 ms
7,492 KB
testcase_02 AC 4 ms
7,620 KB
testcase_03 AC 11 ms
7,748 KB
testcase_04 AC 10 ms
7,624 KB
testcase_05 AC 11 ms
7,488 KB
testcase_06 AC 10 ms
7,492 KB
testcase_07 AC 11 ms
7,620 KB
testcase_08 AC 17 ms
7,496 KB
testcase_09 AC 22 ms
7,620 KB
testcase_10 AC 23 ms
7,496 KB
testcase_11 AC 22 ms
7,624 KB
権限があれば一括ダウンロードができます

ソースコード

diff #

#include <bits/stdc++.h>
using namespace std;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 1000000007;
const ll INF = mod * mod;
const int INF_N = 1e+9;
typedef pair<int, int> P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
typedef pair<ll, ll> LP;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-12;
const ld pi = acos(-1.0);
//typedef vector<vector<ll>> mat;
typedef vector<int> vec;

//繰り返し二乗法
ll mod_pow(ll a, ll n, ll m) {
	ll res = 1;
	while (n) {
		if (n & 1)res = res * a%m;
		a = a * a%m; n >>= 1;
	}
	return res;
}

struct modint {
	ll n;
	modint() :n(0) { ; }
	modint(ll m) :n(m) {
		if (n >= mod)n %= mod;
		else if (n < 0)n = (n%mod + mod) % mod;
	}
	operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint &a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }
modint operator-=(modint &a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }
modint operator*=(modint &a, modint b) { a.n = ((ll)a.n*b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, int n) {
	if (n == 0)return modint(1);
	modint res = (a*a) ^ (n / 2);
	if (n % 2)res = res * a;
	return res;
}

//逆元(Eucledean algorithm)
ll inv(ll a, ll p) {
	return (a == 1 ? 1 : (1 - p * inv(p%a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }

const int max_n = 1 << 18;
modint fact[max_n], factinv[max_n];
void init_f() {
	fact[0] = modint(1);
	for (int i = 0; i < max_n - 1; i++) {
		fact[i + 1] = fact[i] * modint(i + 1);
	}
	factinv[max_n - 1] = modint(1) / fact[max_n - 1];
	for (int i = max_n - 2; i >= 0; i--) {
		factinv[i] = factinv[i + 1] * modint(i + 1);
	}
}
modint comb(int a, int b) {
	if (a < 0 || b < 0 || a < b)return 0;
	return fact[a] * factinv[b] * factinv[a - b];
}
using mP = pair<modint, modint>;

int dx[4] = { 0,1,0,-1 };
int dy[4] = { 1,0,-1,0 };




void solve() {
    int t; cin >> t;
    vector<ll> ans(t);
    rep(i, t){
        ll a, b, c, x, y, z; cin >> a >> b >> c >> x >> y >> z;
        ll res = INF;
        ll tmp = 0;

        if(a == c){
            if(x <= z){
                a--;
                tmp += x;
            }else{
                c--;
                tmp += z;
            }
        }
        if(a <= 0 || c <= 0){
            ans[i] = -1;
            continue;
        }

        if(b > 2){
            ll tmp1 = tmp;
            if(a > b-1) tmp1 += (a-b+1)*x;
            if(c > b-2) tmp1 += (c-b+2)*z;
            res = min(res, tmp1);

            ll tmp2 = tmp;
            if(a > b-2) tmp2 += (a-b+2)*x;
            if(c > b-1) tmp2 += (c-b+1)*z;
            res = min(res, tmp2);
        }

        ll ma = max(a, max(b, c));
        ll mi = min(a, min(b, c));
        if(a == b || b == c){
            if(mi-1 > 0) res = min(res, tmp + (b-mi+1)*y);
        }else{
            if(mi == b) res = min(res, tmp);
            else{
                if(mi-1 > 0) res = min(res, tmp + (b-mi+1)*y);
            }
        }
        
        if(res == INF) res = -1;
        ans[i] = res;
    }

    rep(i, t) cout << ans[i] << endl;
}

signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  //cout << fixed << setprecision(10);
  //init_f();
  //init();
  //int t; cin >> t; rep(i, t)solve();
  solve();
//   stop
    return 0;
}
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