Jesus Is Here
Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 65535/102400 K (Java/Others)
Total Submission(s): 257 Accepted Submission(s): 175
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
Source
2015 ACM/ICPC Asia Regional Shenyang Online
题意:求第n项中所有字符‘c‘的坐标差的和。
分析:每一个新的字符串都是由前两个字符串合成的,所以得到这个公式:ans[n]=ans[n-1]+ans[n-2]+X(X表示两个字符串连接后增加的值);假设第n-2项中字符‘c‘的个数为c1,字符‘c‘坐标之和为s1,字符总个数为n1,第n-1项中字符‘c‘的个数为c2,字符‘c‘坐标之和为s2,字符总个数为n2;
1、对于n-1这个串里的每个’c‘所增加的值为n-2中’c‘的反向坐标(就是串长 - 坐标,记为cc ),即c2*(c1*n1-s1);(把c1*n1-s1 拆分成一个一个的c相加,就相当于c1个cc相加)
2、对于n-2这个串里的每个’c‘所增加的值就为n-1中’c‘的坐标,即c1*s2。
#include <iostream> #include <cstdio> #include <cstring> #include <stack> #include <queue> #include <map> #include <set> #include <vector> #include <cmath> #include <algorithm> using namespace std; const double eps = 1e-6; const double pi = acos(-1.0); const int INF = 0x3f3f3f3f; const int MOD = 530600414; #define ll long long #define CL(a) memset(a,0,sizeof(a)) ll ans[300000];//答案 ll c[300000];//c的个数 ll s[300000];//c的坐标和 ll d[300000];//长度 int main() { ans[1]=0; ans[2]=0; ans[3]=1; ans[4]=1; c[4]=1; s[4]=3; d[4]=3; ans[5]=5; c[5]=2; s[5]=7; d[5]=5; ans[6]=16; c[6]=3; s[6]=20; d[6]=8; for(int i=7; i<=201314; i++) { ans[i]=(ans[i-1]+ans[i-2]+(((c[i-2]*d[i-1]-s[i-2])%MOD)*c[i-1])%MOD+(c[i-2]*s[i-1])%MOD)%MOD; c[i]=(c[i-1]+c[i-2])%MOD; s[i]=(s[i-1]+s[i-2]+c[i-1]*d[i-1])%MOD;//第(i-1)个字符串里的c坐标都要加上第(i-2)个串的长度 d[i]=(d[i-1]+d[i-2])%MOD; } int T,n; scanf ("%d",&T); for(int cas=1; cas<=T; cas++) { scanf ("%d",&n); printf ("Case #%d: ",cas); printf ("%lld\n",ans[n]); } return 0; }
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