2017 BJ ICPC 石子合并变种 向量基本功及分类考察

E

模拟

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que; get min
const double eps = 1.0e-10;
const double EPS = 1.0e-4;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
//const int maxn = 3e5 + 10;
const int turn[][] = {{, }, { -, }, {, }, {, -}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
int cat[];
priority_queue<pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > que;
int main()
{
        int m, n, x;
        while (cin >> m >> n >> x)
        {
                int anser1 = m;
                int anser2 = ;
                for (int i = ; i <= n; i++)
                {
                        scanf("%d", &cat[i]);
                        que.push(make_pair(, cat[i]));
                }
                while (!que.empty())
                {
                        pair<int, int> cnt = que.top();
                        if (cnt.first >= x)
                        {
                                que.pop();
                                continue;
                        }
                        else
                        {
                                que.pop();
                                if (anser1 > )
                                {
                                        cnt.first += cnt.second;
                                        if (cnt.first <= x)
                                        {
                                                anser1--;
                                        }
                                        else
                                        {
                                                anser1--;
                                                anser2++;
                                        }
                                        que.push(cnt);
                                }
                        }
                }
                cout << anser1 << " " << anser2 << endl;
        }
}

F

#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que; get min
const double eps = 1.0e-10;
const double EPS = 1.0e-4;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
//const int maxn = 3e5 + 10;
const int turn[][] = {{, }, { , }, { , -}, { -, }};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
char f[][];
char ans[];
char anser[][];
int pop = ;
int dx, dy;
int dir;
int n;
bool check(int x, int y)
{
        if (x > n || x < )
        {
                return false;
        }
        if (y > n || y < )
        {
                return false;
        }
        if (anser[x][y] != '.')
        {
                return false;
        }
        return true;
}
int main()
{
        while (cin >> n)
        {
                for (int i = ; i <= ; i++)
                {
                        for (int j = ; j <= ; j++)
                        {
                                anser[i][j] = '.';
                        }
                }
                dir = ;
                dx = dy = ;
                pop = ;
                for (int i = ; i <= n; i++)
                {
                        scanf("%s", f[i] + );
                }
                for (int i = ; i <= n + ; i++)
                {
                        if (i %  == )
                        {
                                for (int j = i - ; j >= ; j--)
                                {
                                        ans[++pop] = f[j][i - j];
                                }
                        }
                        else
                        {
                                for (int j = ; j <= i - ; j++)
                                {
                                        ans[++pop] = f[j][i - j];
                                }

                        }
                }
                for (int i = n + ; i <=  * n; i++)
                {
                        if (i %  == )
                        {
                                for (int j = n; j >= i - n; j--)
                                {
                                        ans[++pop] = f[j][i - j];
                                }

                        }
                        else
                        {
                                for (int j = i - n; j <= n; j++)
                                {
                                        ans[++pop] = f[j][i - j];
                                }
                        }
                }
//                for (int i = 1; i <= pop; i++)
//                {
//                        cout << ans[i];
//                }
//                cout << endl;
                for (int i = ; i <= pop; i++)
                {
                        //cout<<dx<<" "<<dy<<endl;
                        anser[dx][dy] = ans[i];
                        if (!check(dx + turn[dir][], dy + turn[dir][]))
                        {
                                dir = (dir + ) % ;
                        }
                        dx = dx + turn[dir][];
                        dy = dy + turn[dir][];
                }
                for (int i = ; i <= n; i++)
                {
                        for (int j = ; j <= n; j++)
                        {
                                cout << anser[i][j];
                        }
                        cout << endl;
                }
        }
}

J

石子合并 不过合并的范围由2变为L-R（2<=L<=R<=N） 问你最少的花费是多少