zoukankan      html  css  js  c++  java
  • Codeforces Round #104 (Div. 1) C. Lucky Subsequence

    C. Lucky Subsequence
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

    Petya has sequence a consisting of n integers.

    The subsequence of the sequence a is such subsequence that can be obtained from a by removing zero or more of its elements.

    Two sequences are considered different if index sets of numbers included in them are different. That is, the values ​of the elements ​do not matter in the comparison of subsequences. In particular, any sequence of length n has exactly 2n different subsequences (including an empty subsequence).

    A subsequence is considered lucky if it has a length exactly k and does not contain two identical lucky numbers (unlucky numbers can be repeated any number of times).

    Help Petya find the number of different lucky subsequences of the sequence a. As Petya's parents don't let him play with large numbers, you should print the result modulo prime number 1000000007 (109 + 7).

    Input

    The first line contains two integers n and k (1 ≤ k ≤ n ≤ 105). The next line contains n integers ai (1 ≤ ai ≤ 109) — the sequence a.

    Output

    On the single line print the single number — the answer to the problem modulo prime number 1000000007 (109 + 7).

    Sample test(s)
    input
    3 2
    10 10 10
    output
    3
    input
    4 2
    4 4 7 7
    output
    4
    Note

    In the first sample all 3 subsequences of the needed length are considered lucky.

    In the second sample there are 4 lucky subsequences. For them the sets of indexes equal (the indexation starts from 1): {1, 3}, {1, 4},{2, 3} and {2, 4}.

    题意:在n个数里面选出k个,lucky数不能有相同两个被选出来,问方案数

    把数分成两部分,在lucky num 里面dp 

    dp[i][j] 表示前i 个数里面选j个数方案数dp[i][j] = dp[i-1][j]+dp[i-1][j-1]*num[i] 

    num[i]表示第i个lucky数出现的次数

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<cmath>
    #include<queue>
    #include<vector>
    #include<set>
    #include<stack>
    #include<map>
    #include<ctime>
    #include<bitset>
    #define LL long long
    #define ll long long
    #define INF 0x3f3f3f3f
    #define maxn 100010
    #define eps 1e-6
    #define mod 1000000007
    using namespace std;
    
    LL f[maxn],e[maxn] ;
    LL dp[2500][2500] ;
    int num[2500] ;
    LL pow1(int a,int n)
    {
        if(n==0) return 1 ;
        LL ans=pow1(a,n/2) ;
        ans=ans*ans%mod;
        if(n&1) ans=ans*a%mod;
        return ans;
    }
    LL C(int n ,int k )
    {
        if(k>n) return 0 ;
        if(k==0||n==k) return 1;
        return f[n]*e[k]%mod*e[n-k]%mod ;
    }
    void init()
    {
        f[0]=1;
        e[0]=1;
        for( int i = 1 ; i < maxn ;i++){
            f[i]=f[i-1]*i%mod;
            e[i]=pow1(f[i],mod-2) ;
        }
    }
    map<int,int>mm;
    bool check(int n )
    {
        while(n)
        {
            if(n%10 != 4 && n%10 != 7) return false;
            n /= 10 ;
        }
        return true;
    }
    int main()
    {
        int n,m,i,j;
        int k ,cnt1,cnt2 ,cnt ;
        init();
        LL ans;
        while(scanf("%d%d",&n,&k) != EOF)
        {
            ans=0;
            cnt1=cnt=0;
            memset(num,0,sizeof(num));
            mm.clear();
            for( i = 1 ; i <= n ;i++)
            {
                scanf("%d",&m) ;
                if(check(m)){
                    if(mm[m])num[mm[m]]++ ;
                    else
                    {
                        cnt++;
                        mm[m]=cnt;
                        num[cnt]++;
                    }
                }
                else cnt1++;
            }
            memset(dp,0,sizeof(dp)) ;
            dp[0][0]=1;
            for( i = 1 ; i <= cnt ;i++)
            {
                dp[i][0]=dp[i-1][0];
                for( j = 1 ; j <= i ;j++)
                    dp[i][j]=(dp[i-1][j]+dp[i-1][j-1]*num[i]%mod)%mod;
            }
            for( i = 0 ; i <= cnt&&i<=k;i++)
            {
                ans=(ans+dp[cnt][i]*C(cnt1,k-i)%mod)%mod;
            }
            cout << ans<<endl;
        }
        return 0 ;
    }
    View Code
  • 相关阅读:
    [pixhawk笔记]8-半物理仿真环境
    Python超参数自动搜索模块GridSearchCV上手
    Oriented Response Networks 阅读笔记(一)
    聚类算法评价指标学习笔记
    基于sklearn的常用分类任务指标Python实现
    使用h5py库读写超过内存的大数据
    基于MXNet使用自己的图像数据集训练网络--准备数据与撰写预处理脚本
    在Ubuntu操作系统中添加环境变量
    Jetson TK1 开发板初用体会
    一条脚本搞定OpenCV
  • 原文地址:https://www.cnblogs.com/20120125llcai/p/4077300.html
Copyright © 2011-2022 走看看