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  • hdu 5564 Clarke and digits 矩阵快速幂优化数位dp

    Clarke and digits

    Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)



    Problem Description
    Clarke is a patient with multiple personality disorder. One day, Clarke turned into a researcher, did a research on digits. 
    He wants to know the number of positive integers which have a length in [l,r] and are divisible by 7 and the sum of any adjacent digits can not be k.
     
    Input
    The first line contains an integer T(1T5), the number of the test cases. 
    Each test case contains three integers l,r,k(1lr109,0k18).
     
    Output
    Each test case print a line with a number, the answer modulo 109+7.
     
    Sample Input
    2 1 2 5 2 3 5
     
    Sample Output
    13 125 Hint: At the first sample there are 13 number $7,21,28,35,42,49,56,63,70,77,84,91,98$ satisfied.
     
    Source

    思路:显然数位,dp[ i ][ j ][ pre ]+=dp[ i-1 ][ j1 ][ pre1 ] &&pre1+pre!=k&&0<=k<=9&&(j1%10+pre)%7==j

       但是r太大,考虑矩阵优化;

       

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<string>
    #include<queue>
    #include<algorithm>
    #include<stack>
    #include<cstring>
    #include<vector>
    #include<list>
    #include<bitset>
    #include<set>
    #include<map>
    #include<time.h>
    using namespace std;
    #define LL long long
    #define bug(x)  cout<<"bug"<<x<<endl;
    const int N=5e4+10,M=1e6+10,inf=1e9+10,MOD=1e9+7;
    const LL INF=1e18+10,mod=1e9+7;
    const double eps=(1e-8),pi=(4*atan(1.0));
    
    struct Matrix
    {
        const static int row=71;
        int a[row][row];//矩阵大小根据需求修改
        Matrix()
        {
            memset(a,0,sizeof(a));
        }
        void init()
        {
            for(int i=0;i<row;i++)
                for(int j=0;j<row;j++)
                    a[i][j]=(i==j);
        }
        Matrix operator + (const Matrix &B)const
        {
            Matrix C;
            for(int i=0;i<row;i++)
                for(int j=0;j<row;j++)
                    C.a[i][j]=(a[i][j]+B.a[i][j])%MOD;
            return C;
        }
        Matrix operator * (const Matrix &B)const
        {
            Matrix C;
            for(int i=0;i<row;i++)
                for(int k=0;k<row;k++)
                    for(int j=0;j<row;j++)
                        C.a[i][j]=(C.a[i][j]+1LL*a[i][k]*B.a[k][j])%MOD;
            return C;
        }
        Matrix operator ^ (const int &t)const
        {
            Matrix A=(*this),res;
            res.init();
            int p=t;
            while(p)
            {
                if(p&1)res=res*A;
                A=A*A;
                p>>=1;
            }
            return res;
        }
    };
    Matrix base,one;
    void init(int k)
    {
        memset(base.a,0,sizeof(base.a));
        for(int i=0;i<70;i++)
        {
            int x=(i/10);
            int y=(i%10);
            for(int j=0;j<70;j++)
            {
                int x1=j/10;
                int y1=j%10;
                if((y+x1*10)%7==x&&y1+y!=k)
                    base.a[j][i]=1;
            }
        }
        for(int i=0;i<10;i++)base.a[i][70]=1;
        base.a[70][70]=1;
    }
    int main()
    {
        memset(one.a,0,sizeof(one.a));
        for(int i=1;i<10;i++)one.a[0][(i%7)*10+i]=1;
        int T;
        scanf("%d",&T);
        while(T--)
        {
            int l,r,k;
            scanf("%d%d%d",&l,&r,&k);
            init(k);
            Matrix ansr=one*(base^(r));
            Matrix ansl=one*(base^(l-1));
            LL out=ansr.a[0][70]-ansl.a[0][70];
            out=(out%mod+mod)%mod;
            printf("%lld
    ",out);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/jhz033/p/7616104.html
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