AtCoder Beginner Contest 114 Solution

A 753

Solved.

     #include <bits/stdc++.h>
     using namespace std;
 
     int mp[];
 
     int main()
     {
         mp[] = mp[] = mp[] = ;
         int x; cin >> x;
         puts(mp[x] ? "YES" : "NO");
     }

B 754

Solved.

     #include <bits/stdc++.h>
     using namespace std;
 
     char s[];
 
     int f(int x)
     {
         int res = ;
         for (int i = ; i < ; ++i) res = res *  + s[i + x] - '';
         return res;
     }
 
     int main()
     {
         while (scanf("%s", s + ) != EOF)
         {
             int res = 0x3f3f3f3f, len = strlen(s + );
             for (int i = ; i <= len - ; ++i) res = min(res, abs(f(i) - ));
             printf("%d\n", res);
         }
         return ;
     }

C 755

Solved.

题意：

找出$[1, n]中有多少个只由'7', '5', '3' 组成，并且每个字符至少出现一次的数$

思路：

这样的数不会太多，DFS构造，然后二分

     #include <bits/stdc++.h>
     using namespace std;
 
     vector <int> v;
 
     bool ok(int x)
     {
         int flag[] = {false};
         while (x)
         {
             flag[x % ] = ;
             x /= ;
         }
         if (flag[] ==  || flag[] ==  || flag[] == ) return false;
         return true;
     }
 
     void DFS(int cur, int num)
     {
         if (cur == )
         {
             if (ok(num)) v.push_back(num);
             return;
         }
         DFS(cur + , num);
         DFS(cur + , num *  + );
         DFS(cur + , num *  + );
         DFS(cur + , num *  + );
     }
 
     int main()
     {
         DFS(, );
         sort(v.begin(), v.end());
         v.erase(unique(v.begin(), v.end()), v.end());
         int n;
         while (scanf("%d", &n) != EOF) printf("%d\n", (int)(upper_bound(v.begin(), v.end(), n) - v.begin()));
         return ;
     }

D 756

Upsolved.

题意：

有$N!中所有因子中，有多少因子其拥有的因子个数恰好为75个$

思路：

我们考虑$75 = 75 \cdot 1 = 25 \cdot 3 = 15 \cdot 5 = 5 \cdot 5 \cdot 3$

那么我们处理出$N!中每个质因子一共有多少个，然后考虑质因子个数如何组成因子个数$

考虑一个数$x = a_1^{p_1} \cdot a_2^{p_2} \cdot a_3^{p_3}$

那么$a_1 可以提供的因子个数为 (p_1 + 1) 那么x 的因子个数即 (p_1 \cdot p_2 \cdot p_3)$

然后简单组合一下就可以了

         #include <bits/stdc++.h>
         using namespace std;
 
         int n;
         int cnt[];
         int tot[];
 
         int f(int l, int r)
         {
             int res = ;
             for (int i = l; i <= r; ++i) res += tot[i];
             return res;
         }
 
         int main()
         {
             while (scanf("%d", &n) != EOF)
             {
                 memset(cnt, , sizeof cnt);
                 memset(tot, , sizeof tot);
                 for (int i = ; i <= n; ++i)
                 {
                     int tmp = i;
                     for(int j = ; ; ++j)
                     {
                         while (tmp % j == )
                         {
                             ++cnt[j];
                             tmp /= j;
                         }
                         if (tmp == ) break;
                     }
                 }
                 for (int i = ; i <= ; ++i) ++tot[cnt[i] + ];
                 int res = f(, );
                 res += f(, ) * f(, );
                 res += f(, ) * (f(, ) - );
                 res += f(, ) * f(, );
                 res += f(, ) * (f(, ) - );
                 res += (f(, ) * (f(, ) - ) / ) * f(, );
                 res += ((f(, ) * (f(, ) - ) / ) * (f(, ) - ));
                 printf("%d\n", res);
             }
             return ;
         }