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：...