Jesus Is Here

Jesus Is Here

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/102400 K (Java/Others)
Total Submission(s): 346    Accepted Submission(s): 243

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

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

 

Source

2015 ACM/ICPC Asia Regional Shenyang Online

 题意：s1 = c，s2 = ff，递推，当i>=3,i等于i-2加上i-1构成，即s3 = cff，问第n个字符串中所有c字符的下标差值

第i个字符串的所有c字符的下标差值肯定由i-1的串还有i-2的串的计算过来的，然而怎么计算……需要什么定义什么，让计算机干什么

你想计算下标差值，这么多数，i-1，i-2，定义-数组把所有的下标和存起来，当算i-1和i-2串的差值计算的时候分两步，算i-2串到串末的距离加上i-1串的距离就等于i-1和i-2串差值

 *****************************************************************************************************************************

 ans[i]             =               ans[i-1]+ans[i-2]   +                                       (len[i-2]*num[i-2]-sum[i-2])*num[i-1]                  +                     sum[i-1]*num[i-2];

 i  字符串中所有c字符差值 = i-1和i-2差值之和，再加上i-1中的c和i-2中的c的差值， （= 所有 i-2 中的c的下标到i-2串末尾的距离   +  i-1串的和*i-2中c的数量 

i-2中的c下标到i-2末尾的距离，就等于i-2中c字符 个 总长度减去c字符的下标之和

#include <iostream>
#include<cstdio>
#include<cstring>
#include<math.h>
#include<algorithm>

using namespace std;

const int maxn = 201315;

#define mod 530600414
#define INF 0x3f3f3f3f

long long num[maxn], sum[maxn], len[maxn], ans[maxn];
// num是该下标表示的字符串中c的个数，sum是所有c下标之和，len是字符串长度，ans是所有下标差值之和
int main()
{
    int t, x;
    num[1] = len[1] = sum[1] = 1;
    len[2] = 2;
    sum[2] = num[2] = 0;
    for(int i = 3; i < maxn; i++)
    {
        num[i] = (num[i-1] + num[i-2])%mod;
        len[i] = (len[i-1] + len[i-2])%mod;
        sum[i] = ((sum[i-1] + sum[i-2])%mod + (len[i-2]*num[i-1])%mod)%mod;  //sum表示c下标和，因为i-1串放在了i-2串的后边，所以sum要加上i-1串中的数量*i-2串的长度，就是i串中所有c的下标和
        ans[i] = (ans[i-1]%mod + ans[i-2]%mod + (num[i-2]%mod*len[i-2]%mod - sum[i-2]%mod)%mod*num[i-1]%mod+num[i-2]%mod*sum[i-1]%mod)%mod;
    }
    scanf("%d", &t);
    for(int i = 1; i <= t; i++)
    {
        scanf("%d", &x);
        printf("Case #%d: %I64d
", i, ans[x]);
    }
    return 0;
}