1006 &6685Rikka with Coin

最新推荐文章于 2021-04-23 21:02:27 发布

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本文深入探讨了C++在编程竞赛中的高级应用，包括宏定义优化、位运算技巧、快速输入输出方法、动态规划解决方案等，旨在提高参赛者的算法效率和编程速度。通过实例解析，读者将了解到如何在比赛中快速解决问题，掌握高效的数据结构和算法实现。

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Rikka with Coin

Rikka with Coin

#include <bits/stdc++.h>
#define mem(ar,num) memset(ar,num,sizeof(ar))
#define me(ar) memset(ar,0,sizeof(ar))
#define lowbit(x) (x&(-x))
#define Pb push_back
#define  FI first
#define  SE second
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define IOS ios::sync_with_stdio(false)
#define DEBUG cout<<endl<<"DEBUG"<<endl;
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int    prime = 999983;
const int    INF = 0x7FFFFFFF;
const LL     INFF = 0x7FFFFFFFFFFFFFFF;
const double pi = acos(-1.0);
const double inf = 1e18;
const double eps = 1e-6;
const LL     mod = 1e9 + 7;
LL qpow(LL a, LL b) {LL s = 1; while (b > 0) {if (b & 1)s = s * a % mod; a = a * a % mod; b >>= 1;} return s;}
LL gcd(LL a, LL b) {return b ? gcd(b, a % b) : a;}
int dr[2][4] = {1, -1, 0, 0, 0, 0, -1, 1};
typedef pair<int, int> P;

const int maxn = 100 + 10;
LL w[maxn];
int n;
int dp[6][6][6][21];
LL solve() {
    cin >> n;
    for (int i = 1; i <= n; ++i)
        scanf("%lld", &w[i]);
    for (int i = 1; i <= n; ++i) {
        if (w[i] % 10 != 0) return -1;
        w[i] /= 10;
    }
    LL ans = INF;
    for (int i = 0; i <= 4; ++i) {
        for (int j = 0; j <= 4; ++j) {
            for (int k = 0; k <= 4; ++k) {
                LL t = 0;
                for (int l = 1; l <= n; ++l) {
                    LL t1, t2;
                    t1 = t2 = INF;
                    if (1) {
                        LL tmp = w[l] % 10;
                        // cout << w[l] << " " ;
                        if (!dp[i][j][k][tmp]) {
                            // cout << tmp << endl;
                            t1 = INF;
                        }
                        else {
                            LL sub = (w[l] - tmp) / 10;
                            t1 = sub;
                        }
                    }
                    if (1) {
                        if (w[l] >= 10) {
                            LL tmp = w[l] % 10 + 10;
                            if (!dp[i][j][k][tmp]) {

                                // cout << tmp << endl;
                                t2 = INF;
                            }
                            else {
                                LL sub = (w[l] - tmp) / 10;
                                t2 = sub;
                            }
                        }

                    }

                    t = max(t, min(t1, t2));

                }
                if (t != INF) {
                    // cout << i << " " << j << " " << k << " " << t << endl;
                    ans = min(ans, i + j + k + t);
                }
            }
        }
    }
    return ans;
}
int main(void)
{

    for (int i = 0; i <= 5; ++i) {
        for (int j = 0; j <= 5; ++j) {
            for (int k = 0; k <= 5; ++k) {
                dp[i][j][k][0] = 1;
                for (int l = 20; l > 0; --l) {
                    if (i > 0 && l >= 1)
                        dp[i][j][k][l] |= dp[i - 1][j][k][l - 1];
                    if (j > 0 && l >= 2)
                        dp[i][j][k][l] |= dp[i][j - 1][k][l - 2];
                    if (k > 0 && l >= 5)
                        dp[i][j][k][l] |= dp[i][j][k - 1][l - 5];
                }
            }
        }
    }
    // cout << dp[0][3][1][2] << endl;
    int T; cin >> T;
    while (T--) {
        cout << solve() << endl;
    }


    return 0;
}