AtCoder Beginner Contest 114 Solution

A 753

Solved.

    #include <bits/stdc++.h>
    using namespace std;
     
    int mp[10];
     
    int main()
    {
        mp[7] = mp[5] = mp[3] = 1;
        int x; cin >> x;
        puts(mp[x] ? "YES" : "NO");
    }

B 754

Solved.

    #include <bits/stdc++.h>
    using namespace std;
     
    char s[20];
     
    int f(int x)
    {
        int res = 0;
        for (int i = 0; i < 3; ++i) res = res * 10 + s[i + x] - '0';
        return res;
    }
     
    int main()
    {
        while (scanf("%s", s + 1) != EOF)
        {
            int res = 0x3f3f3f3f, len = strlen(s + 1);
            for (int i = 1; i <= len - 2; ++i) res = min(res, abs(f(i) - 753));
            printf("%d
", res);
        }
        return 0;
    }

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[10] = {false};
        while (x)
        {
            flag[x % 10] = 1;
            x /= 10;
        }
        if (flag[3] == 0 || flag[5] == 0 || flag[7] == 0) return false;
        return true;
    }
     
    void DFS(int cur, int num)
    {
        if (cur == 10) 
        {
            if (ok(num)) v.push_back(num);
            return;
        }
        DFS(cur + 1, num);
        DFS(cur + 1, num * 10 + 3);
        DFS(cur + 1, num * 10 + 5);
        DFS(cur + 1, num * 10 + 7);
    }
     
    int main()
    {
        DFS(0, 0);
        sort(v.begin(), v.end());
        v.erase(unique(v.begin(), v.end()), v.end());
        int n;
        while (scanf("%d", &n) != EOF) printf("%d
", (int)(upper_bound(v.begin(), v.end(), n) - v.begin()));
        return 0;
    }

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[110];
        int tot[2100];
         
        int f(int l, int r)
        {
            int res = 0;
            for (int i = l; i <= r; ++i) res += tot[i];
            return res;
        }
         
        int main()
        {
            while (scanf("%d", &n) != EOF)
            {
                memset(cnt, 0, sizeof cnt); 
                memset(tot, 0, sizeof tot);
                for (int i = 2; i <= n; ++i)
                {
                    int tmp = i;
                    for(int j = 2; ; ++j) 
                    {
                        while (tmp % j == 0) 
                        {
                            ++cnt[j];
                            tmp /= j; 
                        }
                        if (tmp == 1) break; 
                    }
                }
                for (int i = 2; i <= 100; ++i) ++tot[cnt[i] + 1];  
                int res = f(75, 2000);    
                res += f(25, 2000) * f(3, 24);
                res += f(25, 2000) * (f(25, 2000) - 1);
                res += f(15, 2000) * f(5, 14);
                res += f(15, 2000) * (f(15, 2000) - 1);
                res += (f(5, 2000) * (f(5, 2000) - 1) / 2) * f(3, 4);    
                res += ((f(5, 2000) * (f(5, 2000) - 1) / 2) * (f(5, 2000) - 2));
                printf("%d
", res);
            }    
            return 0;
        }