Jesus Is Here（区域赛网络赛选拔）

Jesus Is Here

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/102400 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

I've sent Fang Fang around 201314 text messages in almost 5 years. Why can't she make sense of what I mean?
``But Jesus is here!" the priest intoned. ``Show me your messages."
Fine, the first message is s1=‘‘c" and the second one is s2=‘‘ff".
The i-th message is si=si−2+si−1 afterwards. Let me give you some examples.
s3=‘‘cff", s4=‘‘ffcff" and s5=‘‘cffffcff".

``I found the i-th message's utterly charming," Jesus said.
``Look at the fifth message". s5=‘‘cffffcff" and two ‘‘cff" appear in it.
The distance between the first ‘‘cff" and the second one we said, is 5.
``You are right, my friend," Jesus said. ``Love is patient, love is kind.
It does not envy, it does not boast, it is not proud. It does not dishonor others, it is not self-seeking, it is not easily angered, it keeps no record of wrongs.
Love does not delight in evil but rejoices with the truth.
It always protects, always trusts, always hopes, always perseveres."

Listen - look at him in the eye. I will find you, and count the sum of distance between each two different ‘‘cff" as substrings of the message.

 

Input

An integer T (1≤T≤100), indicating there are T test cases.
Following T lines, each line contain an integer n (3≤n≤201314), as the identifier of message.

 

Output

The output contains exactly T lines.
Each line contains an integer equaling to:

∑i<j:sn[i..i+2]=sn[j..j+2]=‘‘cff"(j−i) mod 530600414,

where sn as a string corresponding to the n-th message.

 

Sample Input

956781131205199312199401201314

 

Sample Output

Case #1: 5Case #2: 16Case #3: 88Case #4: 352Case #5: 318505405Case #6: 391786781Case #7: 133875314Case #8: 83347132Case #9: 16520782
//I have given up the treatment-_-||#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<string>#include<vector>#include<queue>#include<set>#include<map>#include<stack>#include<cmath>#define mst(ss,b) memset((ss),(b),sizeof(ss))#define inf 0x3f3f3f3f#define max1(a,b) (a)>(b)?(a):(b)#define min1(a,b) (a)<(b)?(a):(b)#define LL  __int64#define ll  long long#define lson  l,m,rt<<1#define rson  m+1,r,rt<<1|1#define eps  1e-12using namespace std;const int  N=250005;const int mod=10000000007;long long h[201400][5];int main(){//   h[4]={5,0,0,5,1};//   h[5]={11,5,5,8,2};   h[4][0]=5;   h[4][1]=0;   h[4][2]=0;   h[4][3]=5;   h[4][4]=1;   h[5][0]=11;   h[5][1]=5;   h[5][2]=5;   h[5][3]=8;   h[5][4]=2;   for(int i=6;i<=201315;i++)   {       h[i][4]=(h[i-1][4]%530600414+h[i-2][4]%530600414)%530600414;       h[i][3]=(h[i-1][3]%530600414+h[i-2][3]%530600414)%530600414;       h[i][2]=((h[i-1][4]%530600414)*(h[i-2][1]%530600414)%530600414+(h[i-2][4]%530600414)*(h[i-1][0]%530600414)%530600414+h[i-1][2]%530600414+h[i-2][2]%530600414)%530600414;       //h[i][2]%=530600414;       h[i][0]=(((h[i-1][4]%530600414)*(h[i-2][3]%530600414))%530600414+h[i-2][0]%530600414+h[i-1][0]%530600414)%530600414;       h[i][1]=(((h[i-2][4]%530600414)*(h[i-1][3]%530600414))%530600414+h[i-1][1]%530600414+h[i-2][1]%530600414)%530600414;   }   int T,n,f=1;   scanf("%d",&T);   while(T--)   {       scanf("%d",&n);       printf("Case #%d: %lld\n",f++,h[n][2]);   }   return 0;}