Hdoj ARC of the Dream 4686 "Matrix fast Power"

ARC of DreamTime limit:2000/2000 MS (java/others) Memory limit:65535/65535 K (java/others) Total submission (s): 3126 Accepted Submission (s): 982 Problem Descriptionan ARC of Dream is a curve defined by following function: where a0 = A0 AI = Ai-1*ax+ay B0 = B0 BI = Bi-1*bx+by What is the value of AoD (N) modulo 1,000,000,007? Inputthere is multiple test cases. Process to the End of File. Each test case contains 7 nonnegative integers as follows: N A0 AX AY B0 BX by N is no further than 1018, and all the other integers be no more than 2x109. Outputfor each test case, output AoD (N) modulo 1,000,000,007. Sample Input 11 2 34 5 621 2 34 5 631 2 34 5 6 Sample Output 41341902 Authorzejun Wu (Watashi) Source2013 multi-university Training Contest 9 Recommendzhuyuanchen520 | We have carefully selected several similar problems for you:5390 5389 5388 5387 5386

Test instructions: Look at the formula .... Ai*bi the sum of the first n (0-n-1) of the series.

Problem-Solving ideas: 10^18 the second party. Decisive use of matrix fast power. The key is to find the matrix of multiplication.

1	0	0	1	0
0	Ax	0	0	Ay
0	0	Bx	0	By
0	Ax*by	Bx*ay	Ax*bx	Ay*by
0	0	0	0	1

is to use the matrix above to multiply

Si-1

Ai-1

Bi-1

Ai-1*bi-1

1

The result of multiplying them is

Si

Ai

Bi

Ai*bi

1

Therefore, the first matrix is directly to the N-order fast power. Multiply the second matrix, get the last matrix, and then output the first element.

AC Code:

#include <stdio.h> #include <math.h> #include <vector> #include <queue> #include <string> #include <string.h> #include <stdlib.h> #include <iostream> #include <algorithm>using Namespace Std;typedef long Long ll;const int maxn = 5;const LL modn = 1000000007; LL RES[MAXN][MAXN]; LL A[MAXN][MAXN];    ll A0,ax,ay,b0,bx,by;void Matmul (ll x[maxn][maxn],ll Y[maxn][maxn]) {ll t[maxn][maxn]={0};             for (int i=0;i<maxn;i++) {to (int k=0;k<maxn;k++) if (X[i][k]) for (int j=0;j<maxn;j++)    t[i][j]= (T[I][J]+X[I][K]*Y[K][J]%MODN)%modn; } for (int i=0;i<maxn;i++) for (int j=0;j<maxn;j++) x[i][j]=t[i][j];} void Matrix (LL x[maxn][maxn],ll N) {for (Int. i=0;i<maxn;i++) for (int j=0;j<maxn;j++) res[i][j]= (i==j    );        while (n) {if (n&1) Matmul (res,x);        Matmul (X,X);    n>>=1;    }}void init () {memset (a,0,sizeof (a));    A[0][0]=a[0][3]=1; a[1][1]=ax%MODN;A[1][4]=AY%MODN;    A[2][2]=BX%MODN;A[2][4]=BY%MODN;    A[3][1]=AX*BY%MODN;A[3][2]=BX*AY%MODN;    A[3][3]=AX*BX%MODN;A[3][4]=AY*BY%MODN; A[4][4]=1;}    int main () {LL n,s;        while (scanf ("%lld", &n)!=eof) {LL ans[5];        scanf ("%lld%lld%lld%lld%lld%lld", &a0,&ax,&ay,&b0,&bx,&by);            if (n==0) {printf ("0\n");        Continue        } ans[0]=0;ans[1]=a0;ans[2]=b0;ans[3]=a0*b0%modn;ans[4]=1;        Init ();        Matrix (A,n);        s=0;        for (int i=0;i<maxn;i++) {s= (S+ANS[I]*RES[0][I]%MODN)%modn;    } printf ("%lld\n", S); } return 0;}

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Hdoj ARC of the Dream 4686 "Matrix fast Power"