Blue Bridge Cup Daily Practice

exclusive square

Enumerate six digits, but because he has a strong property, each digit is a different number, then you only need to dfs the full arrangement of 10 digits (only the first 6 digits are taken) and then square the result after It is enough to determine whether the digits have been used in turn.

#include <iostream>

using namespace std;

int st[10];
long long n;

void dfs(int k) {
    
    
  if(k==6) {
    
    
    if(n==203879) return;
    long long cnt=n*n;
    int flag=1;
    while(cnt) {
    
    
      if(st[cnt%10]) {
    
    
        flag=0;
        break;
      }
      cnt/=10;
    }
    if(flag) {
    
    
      cout<<n<<endl;
      exit(0);
    }
    return;
  }
  for(int i=0;i<10;i++) {
    
    
    if(!k&&!i) continue;
    if(!st[i]) {
    
    
        st[i]=1; n=n*10+i;
        dfs(k+1);
        n/=10; st[i]=0;
    }
  }
}

int main()
{
    
    
  dfs(0);
  return 0;
}

---

Amount not available

Complete backpack. There are only two items in total, each item is unlimited, and only needs 01dp (0 means that the value cannot be obtained, 1 means that it can be obtained), and whether each item can be obtained depends on whether or not to take the item Can the number of

#include <iostream>
#include <vector>

using namespace std;

vector<int>res(2);
int q[100010];

int main()
{
    
    
  for(int i=0;i<2;i++) cin>>res[i],q[res[i]]=1;
  for(int i=0;i<2;i++) {
    
    
    for(int j=res[i];j<100010;j++) {
    
    
      q[j]|=q[j-res[i]];
    }
  }
  for(int i=100009;i>=0;i--) {
    
    
    if(!q[i]) {
    
    
      cout<<i<<endl;
      exit(0);
    }
  }
  return 0;
}

---

palindrome date

Enumerate each date, determine whether the date is a legal date, and determine whether the date will ask. It will ask if the date is found and then find it if it is satisfiedABABBABA

#include <iostream>
#include <algorithm>
#include <string>

using namespace std;

int n;
int days[] = {
    
     0,31,28,31,30,31,30,31,31,30,31,30,31 };

string rev(string ch) {
    
    
    reverse(ch.begin(), ch.end());
    return ch;
}

bool check(int k) {
    
    
    string res = to_string(k);
    return res == rev(res);
}

bool check_(int k) {
    
    
    string res = to_string(k);
    int a = res[0], b = res[1];
    if(a==b) return false;
    return res[2] == a && res[5] == a && res[7] == a && res[3] == b && res[4] == b && res[6] == b;
}

bool check_date(int k) {
    
    
    int day = k % 100;
    int month = (k % 10000) / 100;
    int year = k / 10000;
    if (month == 2) {
    
    
        if (year % 400 == 0 || (year % 100 != 0 && year % 4 == 0)) return day <= 28;
        else return day <= 29;
    }
    if(month>12) return false;
    return day <= days[month];
}

int main()
{
    
    
    cin >> n;
    int flag = 0;
    for (int i = n + 1;; i++) {
    
    
        if(!check_date(i)) continue;
        if (!flag && check(i)) {
    
    
            cout << i << endl;
            flag = 1;
        }
        if (check_(i)) {
    
    
            cout << i << endl;
            break;
        }
    }
    return 0;
}

---

joseph ring

The Joseph ring problem is a very classic problem, and the general approach is to find a pattern. But there are more blogs looking for regularities, so I will not write them (actually, I think I am not as good as them). Here is a simulation writing method that can not find regularity in a short time.

#include <iostream>
using namespace std;
int cal(int sum,int n,int num) {
    
    
  if(num==1) return (sum+n-1)%sum;
  return (cal(sum-1,n,num-1)+n)%sum;
}
int main()
{
    
    
  int n,m;
  cin>>n>>m;
  cout<<cal(n,m,n)+1<<endl;
  return 0;
}