Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 195195 Accepted Submission(s): 48940
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 31 2 100 0 0
Sample Output
2
5
解析:求斐波那契的一点改变。n=100000000,一看就知道常规的做法肯定要超时,所以就用到到了矩阵乘法。
代码:
#include<iostream> #include<memory.h> #include<cstdlib> #include<cstdio> #include<cmath> #include<cstring> #include<string> #include<cstdlib> #include<iomanip> #include<vector> #include<list> #include<map> #include<algorithm> typedef long long LL; const int maxn=1000+10; const int mod=7; const int N=2; using namespace std; struct Matrix { int m[N][N]; }; Matrix I= { 1,0, 0,1 }; Matrix multi(Matrix a,Matrix b) { Matrix c; for(int i=0;i<N;i++) { for(int j=0;j<N;j++) { c.m[i][j]=0; for(int k=0;k<N;k++) c.m[i][j]+=a.m[i][k]*b.m[k][j]%mod; c.m[i][j]%=mod; } } return c; } Matrix power(Matrix A,int k) { Matrix ans=I,p=A; while(k>0) { if(k&1)ans=multi(ans,p); p=multi(p,p); k>>=1; } return ans; } int main() { int a,b,n; while(~scanf("%d%d%d",&a,&b,&n)) { Matrix A; if(a==0&&b==0&&n==0)break; A.m[0][0]=a; A.m[0][1]=b; A.m[1][0]=1; A.m[1][1]=0; Matrix ans =power(A,n-2); printf("%d\n",(ans.m[0][0]+ans.m[0][1])%mod); } return 0; }