hdu1005 Number Sequence (思维+规律)

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 158192    Accepted Submission(s): 38749

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

 

题解：因为给的n比较大，直接递归的话很超时，题中给出公式如果直接代入对的话，说明出题人脑子有问题，所以给出了运算公式的数学凡是没有优化的话，超时超内存等等是避免不了的了.

因为对7取余，f(n)<7，所以会出现循坏，也就是f(n)=1&&f(n-1)=1，这是就能知道了，最多也就7*7=49种情况。

#include<cstdio>int f[10005];int main(){long long a,b,n,i;while(scanf("%lld%lld%lld",&a,&b,&n)&&(a||b||n)){f[1]=f[2]=1;for(i=3;i<=1000;i++){f[i]=(a*f[i-1]+b*f[i-2])%7;if(f[i]==1&&f[i-1]==1)  break;}    n=n%(i-2);    f[0]=f[i-2];    printf("%d\n",f[n]);}return 0;}