HDU 1005 F(Contest #1)

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题意: 已知f[1] = f[2] = 1,输入三个数a,b,n,求f[n] = (a*f[n-1]+b*f[n-2])%7的结果分析: f[n-1]和f[n-2]最多为7种情况(0,1,2,3,4,5,6),则它们的组合最多为49,即周期为49. 代码如下: #include <iostream> #include <cstdio> #include <cstring> #include <fstream> #include <ctime>…

HDU 1005 Number Sequence（数列）

HDU 1005 Number Sequence(数列) Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) [Description] [题目描述] 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, an…

HDU 1005（周期问题）

HDU 1005 Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u 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 v…

HDU 2802 F(N)（简单题，找循环解）

题目链接 F(N) Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4579 Accepted Submission(s): 1610 Problem Description Giving the N, can you tell me the answer of F(N)? Input Each test case contains a si…

HDU - 2802 F(N) （周期）

题目链接:HDU 2009-4 Programming Contest 分析:具有一定的周期性——4018处理下就可以A了 Sample Input Sample Output AC代码: #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<iostream> #include<queue> #include<map>…

HDU 1005 Number Sequence(数论)

HDU 1005 Number Sequence(数论) 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). Input The input consists of multipl…

HDU - 1005 Number Sequence 矩阵快速幂

HDU - 1005 Number Sequence 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). Input The input consists of multiple test…