基于C++的拼多多算法在线笔试题示例

#include <cstdio>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
const int N = 1e6 + 10;
long long a[N];
int k;
int partition(int l,int r) {
  while(l != r)
  {
    while(a[r] >= a[l] && r > l)
      r--;
    if(l == r)
      break;
    swap(a[r],a[l]);
    while(a[l] < a[r] && r > l)
      l++;
    if(l < r)
      swap(a[r],a[l]);
  }
  return l;
}
long long solve(int l,int r) {
  int now = partition(l,r);
  if(k < now)
    return solve(l,now-1);
  else if(k > now)
    return solve(now+1,r);
  else
    return a[now];
}
int main() {
  int n;
  while(~scanf("%d", &n)) {
    for(int i = 0; i < n; ++i) {
      scanf("%lld", &a[i]);
    }
    k = n - 1;
   long long x1 = solve(0, n-1);
    k = n - 2;
    long long x2 = solve(0, n-2);
    k = n - 3;
    long long x3 = solve(0, n-3);
    long long Ans = x1 * x2 * x3;
    if(n > 3) {
      k = 0;
      long long y1 = solve(0, n-1);
      k = 1;
      long long y2 = solve(0, n-2);
      Ans = max(Ans, y1*y2*x1);
    }
    printf("%lld\n", Ans);
  }
  return 0;
}

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
const int N = 1e5 + 10;
string c1, c2;
int a[N], b[N], r[N];
void solve(int a[], int b[], int la, int lb) {
  int i, j;
  for(i = 0; i != N; i++) r[i] = 0;
  for(i = 0; i != la; i++)
  {
    for(j = 0; j != lb; j++)
    {
      int k = i + j;
      r[k] += a[i] * b[j];
      while(r[k] > 9)
      {
        r[k + 1] += r[k] / 10;
        r[k] %= 10;
        k++;
      }
    }
  }
  int l = la + lb - 1;
  while(r[l] == 0 && l > 0) l--;
  for(int i = l; i >= 0; i--) cout << r[i];
  cout << endl;
}
int main() {
  while(cin >> c1 >> c2)
  {
    int la = c1.size(), lb = c2.size();
    for(int i = 0; i != la; i++)
      a[i] = (int)(c1[la - i - 1] - '0');
    for(int i = 0; i != lb; i++)
      b[i] = (int)(c2[lb - i - 1] - '0');
    solve(a, b, la, lb);
  }
  return 0;
}

#include <cstdio>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
const int N = 3e6 + 10;
long long w[N], h[N];
int main() {
  int n, m;
  while(~scanf("%d", &n)) {
    for(int i = 0; i < n; ++i) {
      scanf("%lld", &h[i]);
    }
    scanf("%d", &m);
    for(int i = 0; i < m; ++i) {
      scanf("%lld", &w[i]);
    }
    sort(h, h + n);
    sort(w, w + m);
    int Ans = 0;
    for(int i = 0, j = 0; i < n && j < m; ) {
      if(w[j] >= h[i]) {
        ++Ans;
        ++i, ++j;
      }
      else {
        ++j;
      }
    }
    printf("%d\n", Ans);
  }
  return 0;
}

#include <cstdio>
#include <iostream>
#include <string>
#include <queue>
#include <map>
#include <algorithm>
using namespace std;
const int N = 110;
char mz[N][N];
bool vis[N][N][N*10];
int fx[4] = {0, 0, 1, -1};
int fy[4] = {1, -1, 0, 0};
int m, n;
map<char, int> key;
struct node {
  int x, y, cnt, sta;
  node():cnt(0), sta(0) {}
};
queue<node> que;
int bfs(int sx, int sy, int ex, int ey) {
  while(!que.empty()) que.pop();
  node tmp;
  tmp.x = sx, tmp.y = sy;
  que.push(tmp);
  while(!que.empty()) {
    node p = que.front();
    if(p.x == ex && p.y == ey) {
      return p.cnt;
    }
    que.pop();
    for(int i = 0; i < 4; ++i) {
      int newx = p.x + fx[i];
      int newy = p.y + fy[i];
      if(newx < 0 || newx >= m || newy < 0 || newy >= n) continue;
      if(mz[newx][newy] == '0') continue;
      int sta = p.sta;
      if(mz[p.x][p.y] >= 'a' && mz[p.x][p.y] <= 'z') {
        sta |= (1<<key[mz[p.x][p.y]]);
      }
      if(vis[newx][newy][sta]) continue;
      if(mz[newx][newy] >= 'A' && mz[newx][newy] <= 'Z') {
        if((sta & (1<<(key[mz[newx][newy] - 'A' + 'a'])))== 0) {
          continue;
        }
      }
      vis[newx][newy][sta] = true;
      tmp.x = newx, tmp.y = newy, tmp.cnt = p.cnt + 1, tmp.sta = sta;
      que.push(tmp);
    }
  }
  return -1;
}
int main() {
  while(~scanf("%d %d", &m, &n)) {
    int sx, sy, ex, ey;
    int cnt = 0;
    for(int i = 0; i < m; ++i) {
      scanf("%s", mz[i]);
      for(int j = 0; j < n; ++j) {
        if(mz[i][j] == '2') {
          sx = i, sy = j;
        }
        if(mz[i][j] == '3') {
          ex = i, ey = j;
        }
        if(mz[i][j] >= 'a' && mz[i][j] <= 'z') {
          key[mz[i][j]] = cnt++;
        }
      }
    }
    for(int i = 0; i < m; ++i) {
      for(int j = 0; j < n; ++j) {
        for(int s = 0; s < (1<<cnt); ++s) {
          vis[i][j][s] = false;
        }
      }
    }
    int Ans = bfs(sx, sy, ex, ey);
    printf("%d\n", Ans);
  }
  return 0;
}