渣渣ACM日记——1005-Number Sequence（HDOJ）

   Number Sequence
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 124487    Accepted Submission(s): 30244



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 3
1 2 10
0 0 0
 


Sample Output
2

5

#include "iostream"using namespace std;const int MOD=7;int f[1000];int main(){int a,b,n,i;int result;f[1]=1;f[2]=1;while(cin>>a>>b>>n){if(a==0&&b==0&&n==0)break;if(n<=2){cout<<1<<endl;continue;}for(i=3;i<=n;i++){f[i]=(a*f[i-1]+b*f[i-2])%MOD;//cout<<i<<"----"<<f[i]<<endl;if(i>4&&f[i]==f[4]&&f[i-1]==f[3])//这里坑爹的要i>4 找了好久错误 break;}if(i<n){int tmp=(n-2)%(i-2-2);if(tmp!=0)result=f[tmp+2];else result=f[i-2];}elseresult=f[n];cout<<result<<endl;}return 0;}//超时了 很明显！ /*#include "iostream"using namespace std;const int MOD=7;int main(){int a,b,n,i;while(cin>>a>>b>>n){if(a==0&&b==0&&n==0)break;if(n<=2){cout<<1<<endl;continue;}int last1=1,last2=1,now;for(i=3;i<=n;i++){now=(a*last2+b*last1)%MOD;last1=last2;last2=now;//cout<<i<<" ---- "<<now<<endl; //通过观察可以看出 每一行都有属于自己规律 }cout<<now<<endl;}return 0;}*///内存超出  不要用数组 /*#include "iostream"using namespace std;const int MOD=7;int f[100000005];int main(){int a,b,n,i;f[1]=1;f[2]=1;while(cin>>a>>b>>n){if(a==0&&b==0&&n==0)break;for(i=3;i<=n;i++){f[i]=(a*f[i-1]+b*f[i-2])%MOD;}cout<<f[n]<<endl;}return 0;}*/