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\n",num1,f[d]);
    }
    return 0;
}