NYOJ Number Sequence-----Problem F

Number Sequence
时间限制:1000 ms | 内存限制:65535 KB
难度:2
描述
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).
输入
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.
输出
For each test case, print the value of f(n) on a single line.
样例输入
1 1 3
1 2 10
0 0 0

样例输出
2
5来源
ZJCPC2004
上传者
zinber

冰这个贱男出的题,当时他的数据比较水哈,暴力即可过;但是比赛后,他有重新出了数据,重判……给哥哥判下来了!

这题说不上是难题,但是出的很好,需要思维啊!

题解:如题,f[i]一定是0-6中的某个数,那么f[i-1]与f[i-2]的所有可能组合也只有49种,也就是说,任意两个0-6之间的数的组合超过49组后,必须出现重复的!!出现重复的话那就会循环下去了!所以不用再去算了!

View Code
 
#include <stdio.h>
int f[52];
int main()
{
int i,j,a,b,n;
f[1]=f[2]=1;
while(scanf("%d %d %d",&a,&b,&n),n)
{
for(i=3;i<52;i++) f[i]=(a*f[i-1]+b*f[i-2])%7;
if(n<52) printf("%d\n",f[n]);
else
{
for(i=1;i<50;i++)
for(j=i+1;j<51;j++)
if(f[i]==f[j] && f[i+1]==f[j+1]) goto here;
here: printf("%d\n",f[(n-i)%(j-i)+i]);
}
}
return 0;
}



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