杭电ACM1005 C做法

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Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 173466    Accepted Submission(s): 42860


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
 
脑子锈了不少,简单的题目都要想不少时间了。。。参考了其他大神的一些想法,贴个网址
http://blog.csdn.net/middle544291353/article/details/7723427
余数出现的循环周期为7*7=49次,因此我直接将需要输出的n%49,这样只要计算一次49个数,之后出现的到之前数组找就行,有看到大神用long int类型,我用的是int类型也AC了,直接贴代码吧

#include<stdio.h>int main(){int a,b,n;int f[100];f[1] = 1,f[2] = 1;while(scanf("%d%d%d",&a,&b,&n)!=EOF&&(a!=0)&&(b!=0)&&(n!=0)){if(n==1||n==2) printf("1\n");else{n %= 49;for(int i=3;i<=n;i++){f[i] = (a*f[i-1]+b*f[i-2])%7;}printf("%d\n",f[n]);}}return 0;}