HDU 1005 Number Sequence
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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1005
Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 85249 Accepted Submission(s): 20209
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
25
解析:
这是一道寻找循环点的问题,可能很多人在杭电上通过了这个题目,但是我建议大家将自己的代码再贴到另一个OJ上进行测试http://zju.acmclub.com/index.php?app=problem_title&id=1&problem_id=2603。
很多人都认为周期是49,但是给出的解题报告都不是很有说服力。
所以,我们可以寻找循环的开头以及周期,然后输出,这样能够保证正确性,当然一开始的记录数组最好能够相对大一些,不然仍然不能通过测试。
代码:
#include <stdio.h>#include <stdlib.h>#include <string.h>#include <math.h>#define min(a,b) (a<b?a:b)#define max(a,b) (a>b?a:b)#define swap(a,b) {(a)=(a)^(b); (b)=(a)^(b); (a)=(a)^(b);}#define MAXN 65535#define INF 1e9int f[1200];int main(){ int a,b,n; int i, j; int flag, term, temp, begin; while(~scanf("%d%d%d", &a, &b, &n), (a||b||n)){ memset(f, 0, sizeof(f)); f[1]=1; f[2]=1; term = n; flag = 0; for(i=3; i<=n&&!flag; i++){ f[i] = (a*f[i-1]+b*f[i-2])%7; for(j = 2; j<i; j++){ if(f[i]==f[j]&&f[i-1]==f[j-1]){ term = i-j; begin = j-2; flag = 1; break; } } } if(flag) printf("%d\n", f[begin+(n-1-begin)%term+1]); else printf("%d\n", f[n]); } return 0;}
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