hdu 1005 Number Sequence
2016-05-05 15:22
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Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 147906 Accepted Submission(s): 35942
[align=left]Problem Description[/align]
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).
[align=left]Input[/align]
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.
[align=left]Output[/align]
For each test case, print the value of f(n) on a single line.
[align=left]Sample Input[/align]
1 1 3
1 2 10
0 0 0
[align=left]Sample Output[/align]
2
5
[align=left]Author[/align]
CHEN, Shunbao
[align=left]Source[/align]
ZJCPC2004
#include<iostream> #include<stdio.h> using namespace std; int dp[1500]; int main() { int a,b,n; //cout<<maxx<<endl; while(~scanf("%d%d%d",&a,&b,&n)&&!(a==b&&b==n&&a==0)) { int i; dp[0]= dp[1]=dp[2]=1; if(n<=2) printf("%d\n",dp ); else { for(i=3; i<1500; i++) { dp[i]=(dp[i-1]*a+dp[i-2]*b)%7; if(dp[i-2]==1&&dp[i-1]==1&&i!=3) break; } if(n%(i-3)) printf("%d\n",dp[n%(i-3)]); else printf("%d\n",dp[i-3]); } } return 0; }
View Code
这道题的标准做法是用矩阵快速幂。但是一个数不断的mod一个数,这个结果一定是一个周
期函数。所以可以通过找规律得到结果。是一个比较取巧的解法。
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