Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 174253 Accepted Submission(s): 43022
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 3 1 2 10 0 0 0
Sample Output
2 5
Author
CHEN, Shunbao
Source
Recommend
JGShining
这个题要注意7*7=49种情况,这是模的规律。用最简单的方法会产生超时现象。。。
#include<stdio.h>
int main ()
{
int a,b,n;
int xun[120];
xun[1]=xun[2]=1,xun[0]=1;
while(scanf("%d%d%d",&a,&b,&n)&&a|b|n)
{
int shi,mo,k=0;
for(int i=3;i<=n;++i)
{
xun[i]=(a*xun[i-1]+b*xun[i-2])%7;
for(int j=2;j<=i-1;++j)
{
if(xun[i]==xun[j]&&xun[i-1]==xun[j-1])
{
shi=j;
mo=i;
k=1;
break;
}
}
if(k==1)
break;
}
if(k==1)
{
printf("%d\n",xun[shi+(n-mo)%(mo-shi)]);
}
else
printf("%d\n",xun[n]);
}
}