Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 169273 Accepted Submission(s): 41762
Problem Description
A number sequence is defined as follows:
f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.
Given A, B, and n, you are to calculate the value of f(n).
f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.
Given A, B, and n, you are to calculate the value of f(n).
Input
The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.
Output
For each test case, print the value of f(n) on a single line.
Sample Input
1 1 3
1 2 10
0 0 0
Sample Output
2
5
Author
CHEN, Shunbao
Source
题意：给你A B 求f(n)
题解：构造矩阵 矩阵快速幂
1 #include<bits/stdc++.h> 2 #define ll long long 3 #define mod 7 4 using namespace std; 5 int A,B,n; 6 struct node 7 { 8 int a[3][4]; 9 }exm,ans; 10 struct node matrix_mulit(struct node aa,struct node bb) 11 { 12 struct node there; 13 for(int i=0;i<2;i++) 14 { 15 for(int j=0;j<2;j++) 16 { 17 there.a[i][j]=0; 18 for(int k=0;k<2;k++) 19 { 20 there.a[i][j]=(there.a[i][j]+(aa.a[i][k]*bb.a[k][j]%mod))%mod; 21 } 22 } 23 } 24 return there; 25 } 26 int matrix_quick(int gg) 27 { 28 exm.a[0][0]=A;exm.a[0][1]=B; 29 exm.a[1][0]=1;exm.a[1][1]=0; 30 ans.a[0][0]=1; ans.a[0][1]=0; 31 ans.a[1][0]=1; ans.a[1][1]=0; 32 while(gg) 33 { 34 if(gg&1) 35 ans=matrix_mulit(exm,ans); 36 exm=matrix_mulit(exm,exm); 37 gg>>=1; 38 } 39 return ans.a[0][0]; 40 } 41 int main() 42 { 43 while(scanf("%d %d %d",&A,&B,&n)!=EOF) 44 { 45 if(A==0&&B==0&&n==0) 46 break; 47 if(n==1||n==2) 48 printf("1 "); 49 else 50 printf("%d ",matrix_quick(n-2)); 51 } 52 return 0; 53 }