hdu5459 Jesus Is Here

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

这题主要是推公式，记c[i]为c字符的个数，s[i]为c字符的坐标和,n[i]为所有字符的总个数，f[i]表示求的距离差。

那么f[i]=(c[i-2]*n[i-2]-s[i-2])*c[i-1]+c[i-2]*s[i-1];其中c[i-2]*n[i-2]-s[i-2]为所有c前半部分到合并中间的距离和。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define ll long long
#define inf 0x7fffffff
#define mod 530600414
#define maxn 201316
ll f[maxn],c[maxn],s[maxn],n[maxn];


void init()
{
    int i,j;
    c[3]=1;s[3]=1;n[3]=3;f[3]=0;
    c[4]=1;s[4]=3;n[4]=5;f[4]=0;
    for(i=5;i<=201314;i++){
        f[i]=(f[i-1]+f[i-2]+ ( (c[i-2]*n[i-2]-s[i-2])%mod   )*c[i-1]%mod+c[i-2]*s[i-1]%mod  )%mod;
        n[i]=(n[i-1]+n[i-2])%mod;
        c[i]=(c[i-1]+c[i-2])%mod;
        s[i]=( (s[i-2]+ s[i-1])%mod+(n[i-2]*c[i-1])%mod )%mod;
    }
}


int main()
{
    int m,i,j,T,num1=0,d;
    scanf("%d",&T);
    init();
    while(T--)
    {
        scanf("%d",&d);
        num1++;
        printf("Case #%d: %lld
",num1,f[d]);
    }
    return 0;
}