1409221809-hd-Number Sequence

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).

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 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; }