hdu 5459(递推好题)

Jesus Is Here

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/102400 K (Java/Others)
Total Submission(s): 512    Accepted Submission(s): 368

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

 

这个题真的很难想，但是hdu划分其难度只有2，那么肯定要好好总结一下了。

给出前9项

c
0
ff
0
cff
0
ffcff
0
cffffcff
5
ffcffcffffcff
16
cffffcffffcffcffffcff
88
ffcffcffffcffcffffcffffcffcffffcff
352
cffffcffffcffcffffcffffcffcffffcffcffffcffffcffcffffcff
1552

 

题意：si = si-1+si-2 ，问第i个串中所有c距离之和为多少??

设dp[i]代表第i个串中c的距离之和.

那么dp[i] = dp[i-1]+dp[i-2]+Δ

关键是怎么求Δ。

我们设第i个串的长度为 len[i]，那么容易得到 len[i] = len[i-1]+len[i-2] , len[1]=1,len[2]=2;

设第i个串中c的数量为 num[i]，那么也容易得到 num[i]=num[i-2]+num[i-2],num[1]=1,num[2]=0;

接下来是最重要的:我们设dis[i]代表第i个串中所有的c到末尾的距离之和.那么dis[i]=dis[i-1]+dis[i-2]+num[i-2]*len[i-1],dis[1]=0,dis[2]=0,dis[i-1]+dis[i-2]的话不用说,num[i-2]*len[i-1]代表的是第i-2个串的所有的c到第i-1个串末尾的距离之和,那么增量肯定就是加上第i-2个串中所有的c的个数乘上第i-1个串的长度了。

接下来是如何将所有的条件用上了。

分段考虑：

1.考虑将i-2串中所有的c出发"跳跃"到i-2串的末尾,每个c进行num[i-1]次跳跃,所以总距离为dis[i-2]*num[i-1]

2.接下来考虑i-1串所有的c出发跳跃到i-1串的开头，这样是没办法直接求的，我们知道所有的c跳跃到末尾是dis[i-1]，那么假设所有的c跳跃到开头走的总距离是L，那么L+dis[i-1]=len[i-1]*num[i-1]，每个点进行i-2次跳跃，总共的距离是num[i-2]*(len[i-1]*num[i-1]-dis[i-1])

所以最后的结果为dp[i]=dp[i-1]+dp[i-2]+dis[i-2]*num[i-1]+num[i-2]*(len[i-1]*num[i-1]-dis[i-1])

记得碰到-号多加个mod..不然变成负数就错了！

#include<stdio.h>
#include<iostream>
#include<string.h>
#include <stdlib.h>
#include<math.h>
#include<algorithm>
using namespace std;
typedef long long LL;
const int N = ;
const LL mod =;
LL num[N],len[N],dis[N],dp[N];
void init()
{
    num[]=,num[]=,num[]=,num[]=;
    len[]=,len[]=,len[]=,len[]=;
    dis[]=dis[]=,dis[]=dis[]=;
    dp[]=dp[]=dp[]=dp[]=,dp[]=;
    for(int i=;i<=;i++){
        num[i]=(num[i-]+num[i-])%mod;
        len[i]=(len[i-]+len[i-])%mod;
    }
    for(int i=;i<=;i++){
        dis[i]=((dis[i-]+dis[i-])%mod+num[i-]*len[i-]%mod)%mod;
    }
    for(int i=;i<=;i++){
        dp[i] = ((dp[i-]+dp[i-])%mod+num[i-]*dis[i-]%mod+
                 (len[i-1]*num[i-1]%mod-dis[i-1]+mod)%mod*num[i-]%mod)%mod;
    }
}
int main(){
    init();
    int tcase;
    scanf("%d",&tcase);
    for(int i=;i<=tcase;i++){
        int n;
        scanf("%d",&n);
        printf("Case #%d: %lld\n",i,dp[n]);
    }
    return ;
}