hdoj 5459 Jesus Is Here 【求公式 递推】

Jesus Is Here

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/102400 K (Java/Others)
Total Submission(s): 87    Accepted Submission(s): 57

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

 

题意：给你两个初始串s1  = cff， s2 = ffcff。已知s[i] = s[i-2] + s[i-1]即串s[i-2] 后面接上串s[i-1]，问你第N个串中所有字符c之间的距离之和。

一开始没看这道题，看了之后发现很简单。。。

思路：对每个串，我们用结构体Node保留四个信息——

串长度len，串中字符c的个数num，串中所有字符c的位置之和sum，串中所有字符c之间的距离之和ans。

我们可以得到公式

一、node[i].len = node[i-1].len + node[i-2].len;

二、node[i].num = node[i-1].num + node[i-2].num;

三、node[i].sum = node[i-1].sum + node[i-2].sum + node[i-2].len*node[i-1].num。

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

注意取余的操作，要不测试数据都过不了。

AC代码：

#include <cstdio>#include <cstring>#define MAXN 201314+1#define MOD 530600414#define LL long longstruct Node{    LL len, num, sum, ans;};Node node[MAXN];void getNode(){    node[3].len = 3, node[3].num = 1;    node[3].sum = 1, node[3].ans = 0;    node[4].len = 5, node[4].num = 1;    node[4].sum = 3, node[4].ans = 0;    for(int i = 5; i < MAXN; i++)    {        node[i].len = (node[i-1].len + node[i-2].len)%MOD;        node[i].num = (node[i-1].num + node[i-2].num)%MOD;        node[i].sum = (node[i-1].sum + node[i-2].sum+(node[i-2].len*node[i-1].num)%MOD)%MOD;        node[i].ans = (node[i-1].ans + node[i-2].ans+(((node[i-2].len*node[i-2].num-node[i-2].sum)%MOD)*node[i-1].num)%MOD+(node[i-1].sum*node[i-2].num)%MOD)%MOD;    }}int main(){    getNode();    int t, N, k = 1;    scanf("%d", &t);    while(t--)    {        scanf("%d", &N);        printf("Case #%d: %lld\n", k++, node[N].ans);    }    return 0;}