Qbxt question: Yun

Operation

Question:

The number that contains 4, 7 is the lucky number. For a sequence, the length of K sub-sequence is satisfied: it does not contain two or more identical lucky numbers. (It is illegal to appear twice in 4, but it is legal to appear once in 4 and 7 ).

Analysis:

In 1e9, there are only 2 ^ 10 lucky winners, so you can search them all. Then, calculate the number of occurrences of the lucky number in the sequence. Then you can find the valid number of solutions for these lucky numbers. (It is valid that the selected number cannot appear twice or more than once ). DP [I] [J] indicates that J lucky numbers are selected for the I, and then transferred.

The final statistical answer is how many lucky numbers are selected in the enumeration, which can be obtained directly using the above DP array, and then randomly selected in the non-lucky number to form k.

Code:

 1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #include<iostream> 5 #include<cctype> 6 #include<cmath> 7 #include<set> 8 #include<queue> 9 #include<vector>10 #include<map>11 using namespace std;12 typedef long long LL;13 14 int read() {15     int x = 0, f = 1; char ch = getchar(); for (; !isdigit(ch); ch=getchar()) if (ch==‘-‘) f = -1;16     for (; isdigit(ch); ch=getchar()) x = x * 10 + ch - ‘0‘; return x * f;17 }18 19 #define add(x,y) x+=y, x>=mod?x-=mod:x;20 const int N = 100005;21 const int mod = 1e9 + 7;22 23 vector<int> num;24 map<int,int> Id;25 int cnt[N], a[N], dp[2000][2000], fac[N], ifac[N];26 27 void dfs(int x,LL now) {28     if (x == 11 || now > 1e9) return ; 29     num.push_back(now);30     dfs(x + 1, now * 10 + 4);31     dfs(x + 1, now * 10 + 7);    32 }33 34 int ksm(int a,int b) {35     int res = 1;36     while (b) {37         if (b & 1) res = 1ll * res * a % mod;38         a = 1ll * a * a % mod;39         b >>= 1;40     }41     return res;42 }43 44 void init(int n) {45     dfs(1, 0);46     for (int i = 0; i < num.size(); ++i) Id[num[i]] = i + 1;47     fac[0] = 1;48     for (int i = 1; i <= n; ++i) fac[i] = 1ll * fac[i - 1] * i % mod;49     ifac[n] = ksm(fac[n], mod - 2);50     for (int i = n; i >= 1; --i) ifac[i - 1] = 1ll * ifac[i] * i % mod;51 }52 53 int C(int n,int m) {54     return 1ll * fac[n] * ifac[m] % mod * ifac[n - m] % mod;55 }56 57 int main() {58     59     freopen("lucky.in","r",stdin);60     freopen("lucky.out","w",stdout);61     62     init(100000);63     64     int n = read(), k = read(), m = num.size(), tot = n;65     for (int i = 1; i <= n; ++i) {66         a[i] = read();67         if (Id[a[i]]) cnt[Id[a[i]]] ++, tot --;68     }69     70     dp[0][0] = 1;71     for (int i = 1; i <= m; ++i) {72         for (int j = 0; j <= i; ++j) {73             if (j) add(dp[i][j], 1ll * dp[i - 1][j - 1] * cnt[i] % mod);74             add(dp[i][j], dp[i - 1][j]);75         }76     }77     78     LL ans = 0;79     int L = max(0, k - m), R = min(k, tot);80     for (int i = L; i <= R; ++i) {81         add(ans, 1ll * dp[m][k - i] * C(tot, i) % mod);82     }83     84     cout << ans;85     return 0;86 }

 

Qbxt question: Yun