Jesus Is Here

Time Limit: 1 Sec  

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5459

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

9
5
6
7
8
113
1205
199312
199401
201314

Sample Output

Case #1: 3
Case #2: 2
Case #3: 2
Case #4: -1
Case #5: 2
Case #6: 4
Case #7: 1
Case #8: -1

HINT

题意

f1=c,f2=ff,fn = fn-1+fn-2

然后问你每对cff之间的距离和加起来是多少

题解：

暴力找规律，然后类似fib数列一样递推

不断归纳就好了……

代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

const int N=250000;
const long long pr=530600414;
long long  sum[N],pre[N],tot[N],suf[N],s[N],f[N];

int main()
{
    s[5]=3LL;s[6]=4LL;
    pre[5]=5LL;pre[6]=8LL;
    sum[5]=5LL;sum[6]=11LL;
    suf[5]=5LL;suf[6]=8LL;
    tot[5]=5LL;tot[6]=13LL;
    f[5]=5LL;f[6]=16LL;
    for(int i=7;i<N;i++)
    {
        s[i]=(s[i-1]+s[i-2]-1LL)%pr;
        if(i&1)
        {
            pre[i]=(pre[i-2]+pre[i-1]+5)%pr;
            suf[i]=(suf[i-2]+suf[i-1]+5)%pr;
            sum[i]=(sum[i-2]+(s[i-1]-1)*(pre[i-2]+5)+sum[i-1])%pr;
            tot[i]=(tot[i-1]+(s[i-2]-1)*(suf[i-1]+5)+tot[i-2])%pr;
            f[i]=(f[i-2]+f[i-1]+(s[i-1]-1)*(s[i-2]-1)*5+tot[i-2]*(s[i-1]-1)+sum[i-1]*(s[i-2]-1))%pr;
        }
        else
        {
            pre[i]=(pre[i-2]+pre[i-1]+3)%pr;
            suf[i]=(suf[i-2]+suf[i-1]+3)%pr;
            sum[i]=(sum[i-2]+(s[i-1]-1)*(pre[i-2]+3)+sum[i-1])%pr;
            tot[i]=(tot[i-1]+(s[i-2]-1)*(suf[i-1]+3)+tot[i-2])%pr;
            f[i]=(f[i-2]+f[i-1]+(s[i-1]-1)*(s[i-2]-1)*3+tot[i-2]*(s[i-1]-1)+sum[i-1]*(s[i-2]-1))%pr;
        }
       // cout<<i<<" "<<s[i]<<" "<<pre[i]<<" "<<sum[i]<<" "<<tot[i]<<" "<<f[i]<<endl;
    }
    int T;
    long long n;
    scanf("%d",&T);
    for(int cas=1;cas<=T;cas++)
    {
        scanf("%d",&n);
        printf("Case #%d: %I64d\n",cas,f[n]);
    }
}