Eefun Guessing Words
Eefun is currently learning to read. His way of learning is unique, by trying to make every possible substring from a given string. However, when the string becomes longer, he can no longer remember all the substring. His friends want to test Eefun‘s skill by asking questions, given a string S, is there any substring that begins with letter X and ends with letter Y? According to Eefun, each substring contains at least 2 letters. As the given string S is pretty big, Eefun need your help to answer his friend‘s questions
Input
First line of input contain a single string S which was given by his friends.
Second line contain a number N, the number of questions that Eefun‘s friends ask.
Next N lines each consists of 2 characters, X and Y
Output
For each questions, output a single string "YA" if the substring exists, or "TIDAK" otherwise
(YA means yes and TIDAK means no in Indonesian)
Example
Input: HALO4H OL OA OO L
Output: YAYAYATIDAK
Constraints:
- ‘A‘ ≤ X,Y ≤ ‘Z‘
- 1 ≤ |S| ≤ 1.000.000
- 1 ≤ N ≤ 1.000.000
记一下每个字母第一次和最后一次出现的位置就好了
1 #include<cstdio>
2 #include<iostream>
3 #include<cstring>
4 #include<cstdlib>
5 #include<algorithm>
6 #include<cmath>
7 #include<queue>
8 #include<deque>
9 #include<set>
10 #include<map>
11 #include<ctime>
12 #define LL long long
13 #define inf 0x7ffffff
14 #define pa pair<int,int>
15 #define mkp(a,b) make_pair(a,b)
16 #define pi 3.1415926535897932384626433832795028841971
17 #define mod 100007
18 using namespace std;
19 inline LL read()
20 {
21 LL x=0,f=1;char ch=getchar();
22 while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
23 while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
24 return x*f;
25 }
26 int fst[210],lst[210];
27 char s[1000010],x,y;
28 int n,q;
29 int main()
30 {
31 while (~scanf("%s",s+1))
32 {
33 memset(fst,0,sizeof(fst));
34 memset(lst,0,sizeof(lst));
35 n=strlen(s+1);
36 for (int i=1;i<=n;i++)
37 {
38 if (!fst[s[i]])fst[s[i]]=i;
39 lst[s[i]]=i;
40 }
41 q=read();
42 for (int i=1;i<=q;i++)
43 {
44 scanf(" %c %c",&x,&y);
45 if (fst[x]&&fst[y]&&fst[x]<lst[y])puts("YA");
46 else puts("TIDAK");
47 }
48 }
49 }
Spoj NPC2015A