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.
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.
9
5
6
7
8
113
1205
199312
199401
201314
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
wange2014 | We have carefully selected several similar problems for you:
6343 6342 6341 6340 6339
1 #include <iostream>
2 #include <cstring>
3 #include <cstdio>
4 #include <vector>
5 #include <cmath>
6 #include <algorithm>
7 using namespace std;
8 #define N 201314
9 #define mod 530600414
10 #define gep(i,a,b) for(int i=a;i<=b;i++)
11 #define mem(a,b) memset(a,b,sizeof(a))
12 #define ll long long
13 int t,id;
14 ll f[N+9],c[N+9],s[N+9],n[N+9];
15 /*
16 f ； 任意两个c的坐标差之和
17 c : 字符串里c的个数
18 s : 字符串里所有的c的坐标和
19 n ； 字符串的长度
20 例如 ：
21 cffffcff ffcffcffffcff
22 ((8-6)+(8-1))*3
23 3*2+6*2+11*2
24 上面两个式子的和==f[7]
25 (1+6)+(3+6+11)+8*3==s[7]
26 */
27
28 void init()
29 {
30 c[3]=1,s[3]=1,n[3]=3,f[3]=0;
31 c[4]=1,s[4]=3,n[4]=5,f[4]=0;
32 gep(i,5,N){
33 f[i]=( (f[i-1]+f[i-2])%mod + (((c[i-2]*n[i-2]-s[i-2])%mod+mod)%mod)*c[i-1]%mod +(c[i-2]*s[i-1]%mod) )%mod;
34 c[i]=(c[i-1]+c[i-2])%mod;
35 n[i]=(n[i-1]+n[i-2])%mod;
36 s[i]=((s[i-1]+s[i-2])%mod+(c[i-1]*n[i-2])%mod)%mod;
37 //if(i<=12)printf("%lld %lld %lld %lld\n",f[i],c[i],n[i],s[i]);
38 }
39 }
40 int main()
41 {
42 init();
43 scanf("%d",&t);
44 gep(i,1,t){
45 scanf("%d",&id);
46 printf("Case #%d: %lld\n",i,f[id]);
47 }
48 return 0;
49 }
