第十四届蓝桥杯三月真题刷题训练——第 5 天（3.8）

[Cauchy]

#include <iostream>
using namespace std;
int main()
{
  int n = 2019;
  int res = 0;
  auto ok = [](int x) { while (x) { if (x % 10 == 2 || x % 10 == 4) return false; x /= 10; } return true; };
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      int k = n - i - j; 
      if (k >= j + 1 && ok(i) && ok(j) && ok(k)) {
        res++;
      }
    }
  }
  cout << res;
  return 0;
}

#include <iostream>
using namespace std;
int main()
{
  for (int i = 1950; i <= 2010; i++) {
    for (int j = 1; j <= 12; j++) {
      for (int k = 1; k <= 31; k++) {
        int x = (i * 100 + j) * 100 + k;
        if (x % 2012 == 0 && x % 3 == 0 && x % 12 == 0 && j == 6) {
          cout << x << endl;
          return 0;
        }
      }
    }
  }
  return 0;
}

#include <iostream>
using namespace std;
int main()
{
  int n; cin >> n;
  int x = 0, y = 0;
  int a[n]; for (auto &c: a) {cin >> c; if(c >= 85) x++; else if(c >= 60) y++; }
  y += x;
  printf("%.0lf%%\n%.0lf%%\n", 1.0 * y / n * 100, 1.0 * x / n * 100);
  return 0;
}

#include <iostream>
#include <cstring>
using namespace std;
int main()
{
  int n; cin >> n;
  int a[n + 1]; for (int i = 1; i <= n; i++) cin >> a[i];
  
  int prime[n + 1]; memset(prime, 0x3f, sizeof prime);
  auto init = [&](){
  	for (int i = 2; i <= n; i++) 
  		for (int j = 1; j <= n / i; j++)
  			prime[i * j] = min(prime[i * j], i);
  			
  };
  init();
  // for (int i = 1; i <= n; i++) cout << prime[i] << ' '; cout << endl;
  
  int f[n + 1]; memset(f, -0x3f, sizeof f);
  f[1] = a[1];
  for (int i = 1; i <= n; i++) {
    int j = prime[n - i];
	if (j == 0x3f3f3f3f) j = 1;
    for (int k = i + 1; k <= n && k <= i + j; k++) {
      f[k] = max(f[k], f[i] + a[k]);
    }
  }
  // for (int i = 1; i <= n; i++) cout << f[i] << ' '; cout << endl;
  cout << f[n] << endl;
  return 0;
}

#include <iostream>
#include <cstring>
using namespace std;
int main()
{
  int n; cin >> n;
  int a[n + 1]; for (int i = 1; i <= n; i++) cin >> a[i];
  int f[n + 1]; memset(f, -0x3f, sizeof f);
  f[1] = a[1];
  for (int i = 1; i <= n; i++) {
    int x = n - i;
    int j;
    for (j = 2; j <= x; j++) if (x % j == 0) break;
    if (x % j != 0) j = 1; 
    for (int k = i + 1; k <= n && k <= i + j; k++) {
      f[k] = max(f[k], f[i] + a[k]);
    }
  }
  cout << f[n] << endl;
  return 0;
}