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)



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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).

[align=left]Output[/align]
Each test case print a line with a number, the answer modulo 109+7.

[align=left]Sample Input[/align]

2
1 2 5
2 3 5

[align=left]Sample Output[/align]

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.

[align=left]Source[/align]
BestCoder Round #62 (div.2)
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\n",out);
}
return 0;
}