Description:
Didi is a curious baby. One day, she finds a curious game, which named Rock Paper Scissors Lizard Spock.
The game is an upgraded version of the game named Rock, Paper, Scissors. Each player chooses an option . And then those players show their choices that was previously hidden at the same time. If the winner defeats the others, she gets a point.
The rules are as follows.
Scissors cuts Paper
Paper covers Rock
Rock crushes Lizard
Lizard poisons Spock
Spock smashes Scissors
Scissors decapitates Lizard
Lizard eats Paper
Paper disproves Spock
Spock vaporizes Rock
(and as it always has) Rock crushes Scissors.
(this pic is from baike.baidu.com)
But
Didi is a little silly, she always loses the game. In order to keep her
calm, her friend Tangtang writes down the order on a list and show it
to her. Didi also writes down her order on another list, like 
.
(Rock-R Paper-P Scissors-S Lizard-L Spock-K)
However, Didi may skip some her friends' choices to find the position to get the most winning points of the game, like 
Can you help Didi find the max points she can get?
Input:
The first line contains the list of choices of Didi's friend, the second line contains the list of choices of Didi.
(1<=len(s2)<=len(s1)<=1e6)
Output:
One line contains an integer indicating the maximum number of wining point.
输出时每行末尾的多余空格，不影响答案正确性
样例输入
RRRRRRRRRLLL
RRRS
样例输出
3
发现这题就是个套路水题
字符集很小，枚举计算每个字符对每个开始位置作出的贡献就行了
就很套路 qaq
1 #include"bits/stdc++.h"
2 #define sd(x) scanf("%lf",&(x));
3 #define sld(x) scanf("%lld",&(x));
4 using namespace std;
5
6 const int maxn = 2e6+10;
7 const double Pi = acos(-1.0);
8
9 struct cp
10 {
11 double x,y;
12 cp (double xx=0,double yy=0)
13 {x=xx,y=yy;}
14 }a[maxn],b[maxn];
15 cp operator + (cp a,cp b){ return cp(a.x+b.x , a.y+b.y);}
16 cp operator - (cp a,cp b){ return cp(a.x-b.x , a.y-b.y);}
17 cp operator * (cp a,cp b){ return cp(a.x*b.x-a.y*b.y , a.x*b.y+a.y*b.x);}//不懂的看复数的运算那部分
18
19 int n,m;
20 int l,r[maxn];
21 int limit = 1;
22
23
24 inline void fft(cp *a,int ff)
25 {
26 for(int i=0;i<limit;i++)
27 if(i<r[i])swap(a[i],a[r[i]]);
28 for(int mid=1;mid<limit;mid<<=1)
29 {
30 cp wn(cos(Pi/mid) , ff*sin(Pi/mid));
31 for(int R=mid<<1,j=0;j<limit;j+=R)
32 {
33 cp w(1,0);
34 for(int k=0;k<mid;k++,w=w*wn)
35 {
36 cp x=a[j+k],y=w*a[j+mid+k];
37 a[j+k]=x+y;
38 a[j+mid+k]=x-y;
39 }
40 }
41
42 }
43
44
45 }
46 char s[2000000];
47 char t[2000000];
48 int ans[2000000];
49
50 int main()
51 {
52
53 cin>>s>>t;
54 n=strlen(s); m=strlen(t);
55 reverse(t,t+m);
56 // cout<<t<<endl;
57 string mm="SPRLKSPRLK";
58 while(limit<=n+m)limit<<=1,l++;
59 for(int i=0;i<limit;i++)
60 r[i]=(r[i>>1]>>1)|( (i&1)<<(l-1));
61
62 for(int i=0;i<5;i++)
63 {
64 // cout<<mm[i]<<endl;
65 for(int j=0;j<limit;j++)a[j].x=a[j].y=b[j].x=b[j].y=0;
66 for(int j=0;j<m;j++)a[j].x=(t[j]==mm[i]);
67 for(int j=0;j<n;j++)b[j].x=(s[j]==mm[i+1] || s[j]==mm[i+3]);
68 //for(int j=0;j<m;j++)cout<<a[j].x<<" ";puts("");
69 // for(int j=0;j<n;j++)cout<<b[j].x<<" ";puts("");
70 fft(a,1); fft(b,1);
71 for(int j=0;j<limit;j++)a[j]=a[j]*b[j];
72 fft(a,-1);
73 for(int j=m-1;j<n;j++)
74 ans[j] += int(a[j].x/limit + 0.5);
75 // for(int j=m-1;j<n;j++)cout<<ans[j]<<" "; puts("");
76
77 }
78 int mx=0;
79 for(int j=m-1;j<n;j++)mx=max(mx,ans[j]);
80 cout<<mx;
81
82
83
84
85
86
87
88 }