Team Contests Warmup（2016年多校热身赛，2016年黑龙江省赛）

Team Contests - Warmup

A

题意：...

思路：不会

代码：...

随机 B

题意：给n个点，问是否有一个圆上有最少n/3个点

思路：随机大法好。

代码：...

递推 C

题意：对自然数列{1,2,3,4,5,6...}进行n次操作，每次操作有一个x，意思是前x个保留，后x个删去，再x个保留，后x个删去。。。形成新的序列，问n次操作后第n个数字。

思路：（bh）我开始想二分答案，但是对于n次过程中就被删去的点就无法处理了。那么倒过来做就很方便，已知最后的位置是第n个，那么在第n次操作前的位置应该是，所以倒过来就做好了。结果可能爆long long，用Java写大数。

关键：逆序

代码：

import java.math.*;
import java.io.*;
import java.util.*;

public class Main {
	static int[] a;
	public static void main(String[] args) {
		Scanner in = new Scanner (new BufferedInputStream (System.in));
		int T = in.nextInt ();
		for (int cas=1; cas<=T; ++cas) {
			int n = in.nextInt ();
			a = new int[10005];
			for (int i=1; i<=n; ++i) {
				a[i] = in.nextInt ();
			}
			BigInteger ans = BigInteger.valueOf(n);
			for (int i=n; i>=1; --i) {
				int f = 0;
				if(ans.mod(BigInteger.valueOf(a[i])).compareTo(BigInteger.ZERO) == 0) {
					f = 1;
				}
				BigInteger tmp = ans.divide(BigInteger.valueOf(a[i])).subtract(BigInteger.valueOf(f)).multiply(BigInteger.valueOf(a[i]));
				ans = ans.add(tmp);
			}
			System.out.println (ans);
		}
	}
}

模拟 D

题意：炉石游戏

思路：坑。。。

代码：...

E

题意：麻将

思路：...

代码：...

离散化 F

题意：平面上有n个点，问有多少个不同位置的矩形，满足矩形四条边上的都正好有m个点。

思路：（bh）对点进行坐标离散化点归类，先从左上角开始找到在X容器里，按照y坐标排序后的前(m-1)个点，也就是左下角，同理可以找到右上角，然后从左下角和右上角出发去找右下角，看看是否重合，如果重合表示满足条件，计数即可。

代码：

ans1：

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

const int N = 1e5 + 5;
int n, m;

struct Point {
    int x, y, id;
}p[N];

struct Node {
    int v, id;
    bool operator < (const Node &rhs) const {
        return v < rhs.v;
    }
};

std::vector<Node> X[N], Y[N];
int xs[N], ys[N];
int totx, toty;

int solve() {
    sort (xs, xs+totx);
    totx = unique (xs, xs+totx) - xs;
    sort (ys, ys+toty);
    toty = unique (ys, ys+toty) - ys;
    for (int i=0; i<totx; ++i) {
        X[i].clear ();
    }
    for (int i=0; i<toty; ++i) {
        Y[i].clear ();
    }
    for (int i=1; i<=n; ++i) {
        int px = lower_bound (xs, xs+totx, p[i].x) - xs;
        int py = lower_bound (ys, ys+toty, p[i].y) - ys;
        X[px].push_back ((Node) {p[i].y, p[i].id});
        Y[py].push_back ((Node) {p[i].x, p[i].id});
    }
    for (int i=0; i<totx; ++i) {
        sort (X[i].begin (), X[i].end ());
    }
    for (int i=0; i<toty; ++i) {
        sort (Y[i].begin (), Y[i].end ());
    }
    
    int ret = 0;
    for (int i=1; i<=n; ++i) {
        Node tmp1 = (Node) {p[i].y, p[i].id};
        int px = lower_bound (xs, xs+totx, p[i].x) - xs;
        int vpx = lower_bound (X[px].begin (), X[px].end (), tmp1) - X[px].begin ();
        if (vpx - m + 1 < 0) continue;
        int vpx1 = vpx - m + 1;

        int ty = X[px][vpx1].v;  //(p[i].x, ty)
        Node tmp3 = (Node) {p[i].x, p[i].id};
        int pyt = lower_bound (ys, ys+toty, ty) - ys;
        int vpyt = lower_bound (Y[pyt].begin (), Y[pyt].end (), tmp3) - Y[pyt].begin ();
        if (vpyt + m - 1 >= Y[pyt].size ()) continue;
        int vpyt2 = vpyt + m - 1;

        int x = Y[pyt][vpyt2].v, y = ty;

        Node tmp2 = (Node) {p[i].x, p[i].id};
        int py = lower_bound (ys, ys+toty, p[i].y) - ys;
        int vpy = lower_bound (Y[py].begin (), Y[py].end (), tmp2) - Y[py].begin ();
        if (vpy + m - 1 >= Y[py].size ()) continue;
        int vpy2 = vpy + m - 1;

        int tx = Y[py][vpy2].v;  //(tx, p[i].y)
        Node tmp4 = (Node) {p[i].y, p[i].id};
        int pxt = lower_bound (xs, xs+totx, tx) - xs; //now
        int vpxt = lower_bound (X[pxt].begin (), X[pxt].end (), tmp4) - X[pxt].begin ();
        if (vpxt - m + 1 < 0) continue;
        int vpxt2 = vpxt - m + 1;

        int x2 = tx, y2 = X[pxt][vpxt2].v;
        
        if (x == x2 && y == y2) {
            ret++;
        }
    }
    return ret;
}

int main() {
    int T;
    scanf ("%d", &T);
    while (T--) {
        scanf ("%d%d", &n, &m);
        totx = toty = 0;
        for (int i=1; i<=n; ++i) {
            scanf ("%d%d", &p[i].x, &p[i].y);
            p[i].id = i;
            xs[totx++] = p[i].x;
            ys[toty++] = p[i].y;
        }
        printf ("%d\n", solve ());
    }
    return 0;
}

ans2（zcj）

#include<bits/stdc++.h>
#define mk make_pair

using namespace std;
typedef long long LL;
const int maxn = 300000 + 1000;
int n, m;
pair<int, int> xx[maxn];
vector<LL> v;
vector<int> x[maxn];
vector<int> y[maxn];
map<pair<int, int>, int> p;
int len;

void solve(){
    int cnt = 0;
    for (int i = 0; i < n; i++){
        if (m <= 1) break;
        int tx = xx[i].first;
        int ty = xx[i].second;
        int lenx = x[tx].size(); //表示里面有几个y
        int leny = y[ty].size();//表示里面有几个x
        if (lenx < m) continue;
        if (leny < m) continue;
        //先找x上方的m-1个，里面放着y
        int py = lower_bound(x[tx].begin(), x[tx].end(), ty) - x[tx].begin();//在x的集合里面，y的位置
        if (py + m > lenx) continue;
        int px = lower_bound(y[ty].begin(), y[ty].end(), tx) - y[ty].begin();//在y这个集合里面，x的位置
        if (px + m > leny) continue;
        int posx = y[ty][px + m - 1];
        int posy = x[tx][py + m - 1];//this?
        if (p[mk(posx, posy)]){
            cnt++;
        }
    }
    printf("%d\n", cnt);
}

int main(){
    int t; cin >> t;
    while (t--){
        scanf("%d%d", &n, &m);
        for (int i = 0; i < n; i++){
            LL a, b;
            scanf("%I64d%I64d", &a, &b);
            xx[i] = mk(a, b);
            v.push_back(a);
            v.push_back(b);
        }
        sort(v.begin(), v.end());
        v.erase(unique(v.begin(), v.end()), v.end());
        int lx = 0, ly = 0;
        for (int i = 0; i < n; i++){
            int tx = xx[i].first = lower_bound(v.begin(), v.end(), xx[i].first) - v.begin();
            int ty = xx[i].second = lower_bound(v.begin(), v.end(), xx[i].second) - v.begin();
            //printf("%d %d\n", xx[i].first, xx[i].second);
            p[mk(tx, ty)] = 1;
            x[tx].push_back(ty);
            y[ty].push_back(tx);
            lx = max(lx, tx);
            ly = max(ly, ty);
        }
        for (int i = 0; i <= lx; i++){
            sort(x[i].begin(), x[i].end());
        }
        for (int i = 0; i <= ly; i++){
            sort(y[i].begin(), y[i].end());
        }
        solve();
        //初始化
        p.clear();
        v.clear();
        for (int i = 0; i <= lx; i++){
            x[i].clear();
        }
        for (int i = 0; i <= ly; i++){
            y[i].clear();
        }
        memset(xx, 0, sizeof(xx));
    }
    return 0;
}

G

题意：问[L, R]有多个数字的所有数位相加和是素数的个数。

思路：（bh）数位DP即可，写太着急，long long写成int。。。

代码：

#include <bits/stdc++.h>

typedef long long ll;

bool is_prime(int x) {
    if (x < 2) return false;
    if (x == 2 || x == 3) return true;
    if (x % 6 != 1 && x % 6 != 5) return false;
    for (int i=5; i*i<=x; i+=6) {
        if (x % i == 0 || x % (i + 2) == 0) return false;
    }
    return x != 1;
}

ll dp[20][10][200];
int bit[20];

ll DFS(int pos, int pre, int sum, bool limit) {
    if (pos == -1) {
        return is_prime (sum);
    }
    ll now = dp[pos][pre][sum];
    if (now != -1 && !limit) {
        return now;
    }
    ll ret = 0;
    int d = limit ? bit[pos] : 9;
    for (int i=0; i<=d; ++i) {
        ret += DFS (pos - 1, i, sum + i, limit && i == d);
    }
    if (!limit) {
        dp[pos][pre][sum] = ret;
    }
    return ret;
}

ll solve(ll n) {
    if (n <= 0) {
        return 0;
    }
    int c = 0;
    for (; n>0; n/=10) bit[c++] = n % 10;
    return DFS (c - 1, 0, 0, true);
}

int main() {
    memset (dp, -1, sizeof (dp));
    int T;
    scanf ("%d", &T);
    while (T--) {
        ll a, b;
        scanf ("%I64d%I64d", &a, &b);
        printf ("%I64d\n", solve (b) - solve (a - 1));
    }
    return 0;
}

　　

H

题意：税收

思路：...

代码：...

I

题意：图论

思路：...

代码：...