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

Input

The first line contains an integer

T(1≤T≤5), the number of the test cases.
Each test case contains three integers

l,r,k(1≤l≤r≤109,0≤k≤18).

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;
}