标签:div linker search lld code ane first 矩阵 can
Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
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; }
hdu 5564 Clarke and digits 矩阵快速幂优化数位dp
标签:div linker search lld code ane first 矩阵 can
原文地址:http://www.cnblogs.com/jhz033/p/7616104.html