hdu 5459 Jesus Is Here （费波纳茨递推）

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 65535/102400 K (Java/Others)
Total Submission(s): 250 Accepted Submission(s): 169

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

113

1205

199312

199401

201314

Sample Output

Case #1: 5

Case #2: 16

Case #3: 88

Case #4: 352

Case #5: 318505405

Case #6: 391786781

Case #7: 133875314

Case #8: 83347132

Case #9: 16520782

Source

2015 ACM/ICPC Asia Regional Shenyang Online

题意：s[i] = s[i-1] + s[i-2], s[1] = c, s[2] = ff;

求s[i]中没两个c之间的距离之和

定义：dp[i]表示i所对应的答案，那么dp[i] = dp[i-1] + dp[i-2] + (s[i-1]中的每个c到s[i-2]中的c的距离)

定义：rd[i]表示s[i]中的每个c到s[i]的末尾的距离

ld[i]表示s[i]中的每个c到s[i]的首位的距离

ls[i]表示s[i]的长度

cnt[i]表示s[i]中c的个数

那么有：

rd[i] = rd[i-1] + rd[i-2] + cnt[i-2] * ls[i-1];

ld[i] = ld[i-1] + ld[i-2] + cnt[i-1] * ls[i-2];

dp[i] = dp[i-1] + dp[i-2] + cnt[i-1] * rd[i-2] + cnt[i-2] * ld[i-1];

#include <bits/stdc++.h>

using namespace std;

const int N = 201413;

const int M = 530600414;

typedef long long ll;

ll dp[N], rd[N], ld[N];

ll ls[N], cnt[N];

void init()

{

    ls[3] = 3; cnt[3] = 1; dp[3] = 0;

    ls[4] = 5; cnt[4] = 1; dp[4] = 0;

    rd[3] = 3; ld[3] = 0;

    rd[4] = 3; ld[4] = 2;

    for(int i = 5; i < N; ++i)

    {

        ls[i] = (ls[i - 1] + ls[i - 2]) % M;

        cnt[i] = (cnt[i - 1] + cnt[i - 2]) % M;

        rd[i] = (rd[i - 2] + rd[i - 1] + cnt[i - 2] * ls[i - 1]) % M;

        ld[i] = (ld[i - 2] + ld[i - 1] + cnt[i - 1] * ls[i - 2]) % M;

        dp[i] = (dp[i - 1] + dp[i - 2] + cnt[i - 1] * rd[i - 2] + cnt[i - 2] * ld[i - 1]) % M;

    }

}

int main()

{

    int _, cas = 1; scanf("%d", &_);

    init();

    while(_ --)

    {

        int n; scanf("%d", &n);

        printf("Case #%d: %lld\n",cas++, dp[n] % M);

    }

    return 0;

}