HDOJ 5459 Jesus Is Here（打表规律）

Jesus Is Here

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/102400 K (Java/Others)
Total Submission(s): 405    Accepted Submission(s): 290

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开始看就感觉不会那么简单，暴力一写果然超时，应该是规律，但是实在是没找到，看到别人的规律，秒懂，哎.....ac代码：
#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#define MAXN 210000#define MOD 530600414using namespace std;long long len[MAXN],pos[MAXN],sum[MAXN],num[MAXN];void db(){    len[5]=8;len[6]=13;    num[5]=2;num[6]=3;    pos[5]=7;pos[6]=20;    sum[5]=5;sum[6]=16;    for(int i=7;i<MAXN;i++)    {        len[i]=(len[i-2]+len[i-1])%MOD;        num[i]=(num[i-2]+num[i-1])%MOD;        pos[i]=(pos[i-1]+pos[i-2]+(len[i-2]*num[i-1])%MOD)%MOD;        sum[i]=(sum[i-1]+sum[i-2]+(((len[i-2]*num[i-2]-pos[i-2]))%MOD*num[i-1])%MOD+(pos[i-1]*num[i-2])%MOD)%MOD;    }}int main(){    db();    int t;    int n;    int cas=0;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        printf("Case #%d: %lld\n",++cas,sum[n]);    }    return 0;}