[弱校联萌2016]2016弱校联盟十一专场10.3

比赛链接：https://www.bnuoj.com/v3/contest_show.php?cid=8504#info

A.找两个数乘积是连续上升并且最大的。

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

const int maxn = 1010;
int n, ret;
int a[maxn];

int main() {
  //freopen("in", "r", stdin);
  while(~scanf("%d", &n)) {
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    ret = -1;
    for(int i = 1; i <= n; i++) {
      for(int j = 1; j < i; j++) {
        int tmp = a[i] * a[j];
        int p = tmp % 10; tmp /= 10;
        int q;
        while(tmp) {
          q = tmp % 10;
          if(q + 1 != p) break;
          tmp /= 10; p = q;
        }
        if(tmp == 0) ret = max(ret, a[i]*a[j]);
      }
    }
    printf("%d
", ret);
  }
  return 0;
}

B.从终点开始bfs，记录各点到终点最短路。

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

typedef struct Pair {
  int x, y;
  Pair() {}
  Pair(int x, int y) : x(x), y(y) {}
}Pair;
const int maxn = 220;
const int inf = 900000000;
const int dx[5] = {0,0,1,-1};
const int dy[5] = {1,-1,0,0};
int n, m;
int sx, sy;
int dp[maxn][maxn];
char G[maxn][maxn];
queue<Pair> q;
int rp, rs;

bool ok(int x, int y) {
  return x>=0&&x<n&&y>=0&&y<m;
}

void bfs() {
  rp = rs = inf;
  while(!q.empty()) {
    Pair p = q.front(); q.pop();
    if(G[p.x][p.y] == '$') rs = min(rs, dp[p.x][p.y]);
    if(G[p.x][p.y] == '@') rp = min(rp, dp[p.x][p.y]);
    for(int i = 0; i < 4; i++) {
      int x = p.x + dx[i];
      int y = p.y + dy[i];
      if(ok(x, y) && G[x][y] != '#' && dp[x][y] == -1) {
        dp[x][y] = dp[p.x][p.y] + 1;
        q.push(Pair(x, y));
      }
    }
  }
}

int main() {
  //freopen("in" , "r", stdin);
  while(~scanf("%d%d",&n,&m)) {
    for(int i = 0; i < n; i++) scanf("%s", G[i]);
    for(int i = 0; i < n; i++)
      for(int j = 0; j < m; j++)
        if(G[i][j] == '%') q.push(Pair(i,j));
    memset(dp, -1, sizeof(dp));
    bfs();
    if(rp < rs) puts("Yes");
    else puts("No");
  }
  return 0;
}

D.n对括号最多需要交换n*(n+1)/2次（)))...(((类似这种摆放情况），那么找最短的满足m*(m+1)/2>=n的，然后把额外需要交换的交换完毕就是结果。

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

int n;
string ret;

int main() {
  //freopen("in", "r", stdin);
  while(~scanf("%d", &n)) {
    ret.clear();
    int m = 1;
    while(m * (m + 1) / 2 < n) m++;
    ret = string(m, ')') + string(m, '(');
    int cur = m;
    for(int i = 0; i < m * (m + 1) / 2 - n; i++) {
      swap(ret[cur-1], ret[cur]);
      cur--;
    }
    cout << ret << endl;
  }
  return 0;
}

E.ULL Hash爆搞。

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

typedef long long LL;
typedef unsigned long long ULL;
const ULL mod = ULL(1e9+7);
const int maxn = 100100;
int n;
vector<int> G[maxn];
map<ULL, LL> h;
map<ULL, LL>::iterator it;

ULL dfs(int p, int u) {
  ULL hs = 0;
  for(int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if(v == p) continue;
    hs += dfs(u, v);
  }
  hs *= mod;
  h[++hs]++;
  return hs;
}

int main() {
  //freopen("in", "r", stdin);
  int u, v;
  while(~scanf("%d",&n)) {
    h.clear();
    for(int i = 1; i <= n; i++) G[i].clear();
    for(int i = 0; i < n-1; i++) {
      scanf("%d%d",&u,&v);
      G[u].push_back(v);
      G[v].push_back(u);
    }
    dfs(-1, 1);
    LL ret = 0;
    for(it = h.begin(); it != h.end(); it++) ret += (it->second * it->second);
    cout << (ret - n) / 2 << endl;
  }
  return 0;
}