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  • hdu5564 Clarke and digits

    Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 144    Accepted Submission(s): 84


    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.
     


    这题先列出dp方程,dp[i][x][ (t*10+x)%7 ]+=dp[i-1][j][t],因为i太大,所以要用矩阵快速幂加速。

    可以构造矩阵【dp[i][0][0],dp[i][1][0],..dp[i][0][6],...dp[i][9][6],sum[i-1] 】*A=【dp[i+1][0][0],dp[i+1][1][0],..dp[i+1][0][6],...dp[i+1][9][6],sum[i]】,其中dp[i][j][k]中j表示的是最后一位的数,k表示的是这个数模7后的余数,为了表示前缀和,我们只要把第71列的前10个变为1,dp[70][70]变为1就行了。



    #include<iostream>
    #include<stdio.h>
    #include<stdlib.h>
    #include<string.h>
    #include<math.h>
    #include<vector>
    #include<map>
    #include<set>
    #include<queue>
    #include<stack>
    #include<string>
    #include<algorithm>
    using namespace std;
    typedef long long ll;
    #define inf 99999999
    #define pi acos(-1.0)
    #define MOD 1000000007
    struct matrix{
        ll n,m,i;
        ll data[72][72];
        void init_danwei(){
            for(i=0;i<n;i++){
                data[i][i]=1;
            }
        }
    };
    
    matrix multi(matrix &a,matrix &b){
        ll i,j,k;
        matrix temp;
        temp.n=a.n;
        temp.m=b.m;
        for(i=0;i<temp.n;i++){
            for(j=0;j<temp.m;j++){
                temp.data[i][j]=0;
            }
        }
        for(i=0;i<a.n;i++){
            for(k=0;k<a.m;k++){
                if(a.data[i][k]>0){
                    for(j=0;j<b.m;j++){
                        temp.data[i][j]=(temp.data[i][j]+(a.data[i][k]*b.data[k][j])%MOD )%MOD;
                    }
                }
            }
        }
        return temp;
    }
    
    matrix fast_mod(matrix &a,ll n){
        matrix ans;
        ans.n=a.n;
        ans.m=a.m;
        memset(ans.data,0,sizeof(ans.data));
        ans.init_danwei();
        while(n>0){
            if(n&1)ans=multi(ans,a);
            a=multi(a,a);
            n>>=1;
        }
        return ans;
    }
    
    int main()
    {
        ll n,k,m,i,j,T,l,r,t,x;
        scanf("%lld",&T);
        while(T--)
        {
            scanf("%lld%lld%lld",&l,&r,&k);
            matrix a;
            a.n=a.m=71;
            memset(a.data,0,sizeof(a.data));
            for(t=0;t<=6;t++){
                for(j=0;j<=9;j++){
                    for(x=0;x<=9;x++){
                        if(x+j!=k){
                            a.data[j+10*t  ][x+(t*10+x)%7*10  ]=1;
                        }
                    }
                }
            }
            for(i=0;i<10;i++){
                a.data[i][70]=1;
            }
            a.data[70][70]=1;
    
    
    
            matrix a1;
            a1.n=a1.m=71;
            memset(a1.data,0,sizeof(a1.data));
            for(i=0;i<=70;i++){
                for(j=0;j<=70;j++){
                    a1.data[i][j]=a.data[i][j];
                }
            }
    
            matrix b;
            b.n=1;b.m=71;
            memset(b.data,0,sizeof(b.data));
            for(j=1;j<=9;j++){
                b.data[0][j+j%7*10]=1;
            }
    
            matrix b1;
            b1.n=1;b1.m=71;
            memset(b1.data,0,sizeof(b1.data));
            for(j=1;j<=9;j++){
                b1.data[0][j+j%7*10]=1;
            }
            ll sum1,sum2;
            matrix ans;
            ans=fast_mod(a,l-1);
            matrix cnt;
            cnt=multi(b,ans);
            sum1=cnt.data[0][70];
    
            matrix ans1;
            ans1=fast_mod(a1,r);
            /*
            for(i=0;i<=70;i++){
                for(j=0;j<=70;j++){
                    printf("%lld ",ans1.data[i][j]);
    
                }
                printf("
    ");
            }
            */
    
    
            matrix cnt1;
            cnt1=multi(b1,ans1);
            sum2=cnt1.data[0][70];
    
            printf("%lld
    ",(sum2-sum1+MOD)%MOD );
        }
        return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/herumw/p/9464594.html
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