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HDU4686 Arc of Dream 矩阵快速幂

2024-08-23 22:47:10   220人阅读

Arc of Dream

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 4246    Accepted Submission(s): 1332

Problem Description
An Arc of Dream is a curve defined by following function:
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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?
 
Input
There are 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 more than 1018, and all the other integers are no more than 2×109.
 
Output
For each test case, output AoD(N) modulo 1,000,000,007.
 
Sample Input
1
1 2 3
4 5 6
2
1 2 3
4 5 6
3
1 2 3
4 5 6
 
Sample Output
4
134
1902
 
题意：已知：
a0 = A0 b0 = B0
ai = ai-1*AX+AY
bi = bi-1*BX+BY
Sn = a0b0+a1*b1+...+a(n-1)*b(n-1);
求 Sn
 
题解：矩阵快速幂求n项和

a[i]*b[i] = (a[i-1]*Ax+Ay)(b[i-1]*Bx+By)
= Ax*Bx*a[i-1]*b[i-1]+Ay*Bx*b[i-1]+Ax*By*a[i-1]+Ay*By
s
[s[n-1],a[n-2]*b[n-2], b[n-2], a[n-2], 1]
A
[1 ,0 ,0 ,0 ,0]
[Ax*Bx ,Ax*Bx ,0 ,0 ,0]
[Ay*Bx ,Ay*Bx ,Bx ,0 ,0]
[Ax*By ,Ax*By ,0 ,Ax ,0]
[Ay*By ,Ay*By ,By ,Ay ,1]//n-1
s
[s[n], a[n-1]*b[n-1], b[n-1], a[n-1], 1]

 
#include<bits/stdc++.h>#define N 5#define mes(x) memset(x, 0, sizeof(x));#define ll long longconst ll mod = 1e9+7;const int MAX = 0x7ffffff;using namespace std;struct mat {    ll a[N][N];      mat() {          memset(a, 0, sizeof(a));      }     mat operator * (mat b) {          mat c;          for (int i = 0; i < N; i++)              for (int j = 0; j < N; j++)                  for (int k = 0; k < N; k++)                      c.a[i][j] = (c.a[i][j] + a[i][k] * b.a[k][j]) % mod;          return c;      }};mat f(mat b, ll m) {    mat c;    for (int i = 0; i < N; i++)          c.a[i][i] = 1;    while (m) {        if (m & 1)            c = c * b;          b = b * b;          m >>= 1;      }    return c;  }int main(){    ll n, A0,Ax, Ay, Bx,By,B0;    mat A, s;    while(~scanf("%lld", &n)){        scanf("%lld%lld%lld%lld%lld%lld", &A0, &Ax, &Ay, &B0, &Bx, &By);        if(n == 0){            printf("0\n");            continue;        }        mes(A.a);        mes(s.a);        s.a[0][0] = s.a[0][1] = (A0%mod*B0%mod)%mod;        s.a[0][2] = B0%mod;        s.a[0][3] = A0%mod;        s.a[0][4] = 1;        A.a[0][0] = 1;         A.a[1][1] = A.a[1][0] = (Ax%mod*Bx%mod)%mod;A.a[2][2] = Bx%mod;        A.a[2][1] = A.a[2][0] = (Ay%mod*Bx%mod)%mod;A.a[3][3] = Ax%mod;        A.a[3][1] = A.a[3][0] = (Ax%mod*By%mod)%mod;        A.a[4][0] = A.a[4][1] = (Ay%mod*By%mod)%mod;        A.a[4][2] = By;        A.a[4][3] = Ay;        A.a[4][4] = 1;        A = f(A,n-1);        s = s*A;        printf("%lld\n", (mod+s.a[0][0])%mod);    } }

 

 

HDU4686 Arc of Dream 矩阵快速幂


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