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hdu 1005 找循环节
Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 111101 Accepted Submission(s): 26982
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
ps:本按正常思路写，发现超时。。然后就输数据。。发现原来还有循环。。也就是说找到相对应的循环节就好
#include "stdio.h"
int main()
{
int a,b,n,f[100],i;
while(~scanf("%d%d%d",&a,&b,&n)&&a&&b&&n)
{
f[1]=1;f[2]=1;
for(i=3;i<100;i++)
{
f[i]=(a*f[i-1]+b*f[i-2])%7;
if(f[i-1]==1&&f[i]==1)
break;
}
i-=2;
if(n%i!=0)
printf("%d\n",f[n%i]);
else
printf("%d\n",f[i]);
}
return 0;
}
int main()
{
int a,b,n,f[100],i;
while(~scanf("%d%d%d",&a,&b,&n)&&a&&b&&n)
{
f[1]=1;f[2]=1;
for(i=3;i<100;i++)
{
f[i]=(a*f[i-1]+b*f[i-2])%7;
if(f[i-1]==1&&f[i]==1)
break;
}
i-=2;
if(n%i!=0)
printf("%d\n",f[n%i]);
else
printf("%d\n",f[i]);
}
return 0;
}
hdu 1005 找循环节
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