1409221809-hd-Number Sequence
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Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1328 Accepted Submission(s): 522
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
题目大意
根据题目给定的公式解决。
解题思路
这道题用到了同余定理,但是如果单纯的运用同余定理进行暴力求解,会超出内存。最好的方法是仔细观察题目给定的公式,为什么要对7取余呢?这么大的数据范围肯定会存在循环节,循环电石什么呢?其实我也不知道怎么找,只知道48是一个循环,,
代码
#include<stdio.h>int f[48];int main(){int a,b,n;int i,j,k;while(scanf("%d%d%d",&a,&b,&n)&&a+b+n){f[1]=f[2]=1;for(i=3;i<=n%48;i++)//要找循环节 {j=f[i-1]*a%7;k=f[i-2]*b%7;f[i]=(j+k)%7;}printf("%d\n",f[n%48]);}return 0;}
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