HDU-6395多校7 Sequence（除法分塊+矩陣快速冪）
阿新 • • 發佈：2018-08-15
review lse %d sca code left define hdu fin
F1=A F2=B Fn=C⋅Fn−2+D⋅Fn−1+⌊Pn⌋
Your job is simple, for each task, you should output Fn
Then, for the next T lines, each line consists of 6 integers, A , B, C, D, P, n.
1≤T≤200≤A,B,C,D≤1091≤P,n≤109
Sequence
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1731 Accepted Submission(s): 656
F1=A F2=B Fn=C⋅Fn−2+D⋅Fn−1+⌊Pn⌋
Your job is simple, for each task, you should output Fn
Input The first line has only one integer T, indicates the number of tasks.
Then, for the next T lines, each line consists of 6 integers, A , B, C, D, P, n.
1≤T≤200≤A,B,C,D≤1091≤P,n≤109
Sample Input 2 3 3 2 1 3 5 3 2 2 2 1 4
Sample Output 36 24
#include<iostream> #include<stdio.h> #include<stdlib.h> #include<string.h> #include<math.h> #include<algorithm> #defineMAX 10 #define INF 0x3f3f3f3f #define MOD 1000000007 using namespace std; typedef long long ll; ll p,q; struct mat{ ll a[MAX][MAX]; }; mat operator *(mat x,mat y) { mat ans; memset(ans.a,0,sizeof(ans.a)); for(int i=1;i<=3;i++){ for(int j=1;j<=3;j++){ for(int k=1;k<=3;k++){ ans.a[i][j]+=x.a[i][k]*y.a[k][j]%MOD; ans.a[i][j]%=MOD; } } } return ans; } mat qMod(ll x,mat a,ll n) { mat t; t.a[1][1]=q;t.a[1][2]=p;t.a[1][3]=x; t.a[2][1]=1;t.a[2][2]=0;t.a[2][3]=0; t.a[3][1]=0;t.a[3][2]=0;t.a[3][3]=1; while(n){ if(n&1) a=t*a; n>>=1; t=t*t; } return a; } int main() { int t,i; ll a1,a2,x,n; scanf("%d",&t); while(t--){ scanf("%lld%lld%lld%lld%lld%lld",&a1,&a2,&p,&q,&x,&n); if(n==1) printf("%lld\n",a1); else if(n==2) printf("%lld\n",a2); else{ mat a; a.a[1][1]=a2;a.a[1][2]=0;a.a[1][3]=0; a.a[2][1]=a1;a.a[2][2]=0;a.a[2][3]=0; a.a[3][1]=1;a.a[3][2]=0;a.a[3][3]=0; if(x>=n){ for(i=3;i<=n;i=x/(x/i)+1){ a=qMod(x/i,a,min(n,x/(x/i))-i+1); } } else{ for(i=3;i<=x;i=x/(x/i)+1){ a=qMod(x/i,a,x/(x/i)-i+1); } a=qMod(0,a,n-max(x,2ll)); } printf("%lld\n",a.a[1][1]); } } return 0; }
HDU-6395多校7 Sequence（除法分塊+矩陣快速冪）