Jesus Is Here HDU - 5459 (斐波那契）

最新推荐文章于 2019-08-27 21:39:02 发布

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本文探讨了一种特殊的文本消息生成规则，并通过算法计算不同子串间的特定距离之和。通过对给定序列的数学分析，文章提供了一段C++代码实现，用以高效解决这一问题。

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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 ii-th message is si=si−2+si−1si=si−2+si−1 afterwards. Let me give you some examples.
s3=”cff”, s4=”ffcff” and s5=”cffffcff”.

“I found the ii-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 55.
“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”‘‘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 TT lines.
Each line contains an integer equaling to:

\sum i < j : s n [i . . i + 2] = s n [j . . j + 2] = ‘ ‘ cff " (j - i) m o d 530600414

$\sum_{i<j:s_n[i..i+2]=s_n[j..j+2]=``\texttt{cff}"}(j-i)~mod~530600414$
where snsn as a string corresponding to the nn-th message.
Sample Input
9
5
6
7
8
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

这题真是觉了，明天写详细思路，先睡了

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#define N 201316
#define mod 530600414
using namespace std;
long long dis[N],cnt[N],sum[N],len[N];
void init()
{
    len[1]=1;
    len[2]=2;
    sum[1]=sum[2]=0;
    cnt[1]=1;
    cnt[2]=0;
    dis[1]=dis[2]=0;
    for(int i=3;i<N;i++)
    {
        dis[i]=(dis[i-1]+dis[i-2]+cnt[i-2]*len[i-1])%mod;
        cnt[i]=(cnt[i-1]+cnt[i-2])%mod;
        len[i]=(len[i-1]+len[i-2])%mod;
        sum[i]=(sum[i-1]+sum[i-2]+dis[i-2]*cnt[i-1]+cnt[i-2]*(cnt[i-1]*len[i-1]%mod-dis[i-1]+mod))%mod;
    }
}
int main()
{
    int t;
    int n;
    int ca=1;
    init();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        printf("Case #%d: %lld\n",ca++,sum[n]);
    }
    return 0;
}