杭电ACM-HDU1005-Number Sequence

题目来自杭电ACM: acm.hdu.edu.cn

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 134355    Accepted Submission(s): 32628

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 31 2 100 0 0

 

Sample Output

25

 

Author

CHEN, Shunbao

 

Source

ZJCPC2004

又是一个和斐波那契数列一样的式子，明显不能使用递归的方法，于是就先使用迭代写了代码如下：

#include <stdio.h>#define N 100000010int f[N];int main(){f[1] = 1;f[2] = 1;int a, b, n, i;scanf("%d%d%d", &a, &b, &n);if (n == 0 && a == 0 && b == 0) {}else {for (i = 3; i <= n; ++i) {f[i] = (a * f[i-1] + b * f[i-2]) % 7;}printf("%d", f[n]);}return 0;}

提交后是：Memory Limit Exceeded，内存超出，因此也不能使用迭代的方法。那么此时要么从给出的递归公式推出相应的通项公式，要么从其中寻找一定的规律。因为A，B是不确定值，所以求通项肯定很困难，同时，我们可以看到在公式中有取余的存在，因此数列肯定是循环数列，又因为取7的余数，则数列的每一项一定不大于7，那么最大有7x7=49中可能，于是我们便可以把数组的大小减小到60，在前60个数中找到循环位置，那么便可以轻松求解。

改进后代码如下：

#include <stdio.h>int f[60];int main(){int a, b, n, i, j;while (scanf("%d%d%d", &a, &b, &n) != EOF) {int beg = 1, end = 1;f[0] = f[1] = f[2] = 1;if (n == 0 && a == 0 & b == 0) {break;}for (i = 3; i < 60; ++i) {f[i] = (a*f[i-1] + b*f[i-2]) % 7;for (j = end+2; j < i; ++j) {if (f[i] == f[j] && f[i-1] == f[j-1]) {beg = j - 2;end = i - 2;break;}}}f[0] = f[end-1];f[1] = f[end];if (n == 1 || n == 2) {printf("%d\n", 1);}else{printf("%d\n", f[n%(end-beg)]);}}return 0;}

开始一直wa，后来发现没有在输入一次数据后没有对f[1],f[2]初始化，导致下次输出的结果有误，把数组的初始化放入while循环中就可以ac了。