hdu 1005(number sequence)
题目:http://acm.hdu.edu.cn/showproblem.php?pid=1005
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 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
理解:f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7 如果直接用循环算会造成OVER STACK,所以要因为是对7取余,所以f(n)取值只有可能在0-6之间, f(n - 1)f(n - 2)各有7种可能,所以49次一个循环。
//
// main.c
// hdu1005
//
// Created by zhan_even on 2018/10/19.
// Copyright © 2018年 zhan_even. All rights reserved.
//
#include <stdio.h>
int main(void) {
int a,b,n,i;
int result[50];
result[1] = result[2] = 1;
while (scanf("%d %d %d",&a,&b,&n)!=EOF ) {
if(a == 0 && b==0 && n==0)
break;
else{
for(i=3;i<50;i++){
result[i] = (a*result[i-1]+b*result[i-2])%7;
}
n = n%49;
printf("%d\n",result[n]);
}
}
return 0;
}
也可以利用代码来计算周期,如果又出现 f(n - 1) = f(n ) = 1,则出现循环
//
// main.cpp
// hdu1005way2
//
// Created by zhan_even on 2018/10/19.
// Copyright © 2018年 zhan_even. All rights reserved.
//
#include <iostream>
using namespace std;
int main(int argc, const char * argv[]) {
int result[1000];
int n,a,b;
while (cin>>a>>b>>n) {
if(a==0&&b==0&&n==0)
break;
else{
int i;
for(i=3;i<1000;i++){
result[i] = (a*result[i-1]+b*result[i-2])%7;
if(result[i]==1&&result[i-1]==1)
break;
}
n = n % (i-2);
cout<<result[n]<<endl;
}
}
return 0;
}