BestCoder Round #88

A、Find Q

Accepts: 392
Submissions: 780
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 262144/131072 K (Java/Others)
Problem Description
 

Byteasar is addicted to the English letter 'q'. Now he comes across a string SSS consisting of lowercase English letters.

He wants to find all the continous substrings of SSS, which only contain the letter 'q'. But this string is really really long, so could you please write a program to help him?

Input

The first line of the input contains an integer T(1≤T≤10)T(1leq Tleq10)T(1≤T≤10), denoting the number of test cases.

In each test case, there is a string SSS, it is guaranteed that SSS only contains lowercase letters and the length of SSS is no more than 100000100000100000.

Output

For each test case, print a line with an integer, denoting the number of continous substrings of SSS, which only contain the letter 'q'.

Sample Input
2
qoder
quailtyqqq
Sample Output
1
7
 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <cmath>
 7 #include <time.h>
 8 #include <string>
 9 #include <map>
10 #include <stack>
11 #include <vector>
12 #include <set>
13 #include <queue>
14 #define inf 0x7fffffff
15 #define mod 10000
16 #define met(a,b) memset(a,b,sizeof a)
17 typedef long long ll;
18 using namespace std;
19 const int N = 100;
20 const int M = 100000;
21 const int INF = 0x3f3f3f3f;
22 int x;
23 
24 char s[100002];
25 long long int a[1000];
26 
27 int main()
28 {
29     long long int sum = 0;
30     int t;
31     scanf("%d", &t);
32     while(t--)
33     {
34         scanf("%s", s);
35         int len = strlen(s);
36         long long int cnt = 0;
37         sum = 0;
38         for(int i = 0; i < len; i++)
39         {
40             if(s[i] == 'q')
41             {
42                 cnt++;
43             }else if(cnt > 0){
44                 sum += cnt*(cnt+1)/2;
45                 cnt = 0;
46             }
47         }
48         if(cnt > 0) sum += cnt*(cnt+1)/2;
49         printf("%lld
", sum);
50     }
51 }

Abelian Period

Accepts: 288
Submissions: 984
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 262144/131072 K (Java/Others)
Problem Description

Let SSS be a number string, and occ(S,x)occ(S,x)occ(S,x) means the times that number xxx occurs in SSS.

i.e. S=(1,2,2,1,3),occ(S,1)=2,occ(S,2)=2,occ(S,3)=1S=(1,2,2,1,3),occ(S,1)=2,occ(S,2)=2,occ(S,3)=1S=(1,2,2,1,3),occ(S,1)=2,occ(S,2)=2,occ(S,3)=1.

String u,wu,wu,w are matched if for each number iii, occ(u,i)=occ(w,i)occ(u,i)=occ(w,i)occ(u,i)=occ(w,i) always holds.

i.e. (1,2,2,1,3)≈(1,3,2,1,2)(1,2,2,1,3)approx(1,3,2,1,2)(1,2,2,1,3)≈(1,3,2,1,2).

Let SSS be a string. An integer kkk is a full Abelian period of SSS if SSS can be partitioned into several continous substrings of length kkk, and all of these substrings are matched with each other.

Now given a string SSS, please find all of the numbers kkk that kkk is a full Abelian period of SSS.

Input

The first line of the input contains an integer T(1≤T≤10)T(1leq Tleq10)T(1≤T≤10), denoting the number of test cases.

In each test case, the first line of the input contains an integer n(n≤100000)n(nleq 100000)n(n≤100000), denoting the length of the string.

The second line of the input contains nnn integers S1,S2,S3,...,Sn(1≤Si≤n)S_1,S_2,S_3,...,S_n(1leq S_ileq n)S​1​​,S​2​​,S​3​​,...,S​n​​(1≤S​i​​≤n), denoting the elements of the string.

Output

For each test case, print a line with several integers, denoting all of the number kkk. You should print them in increasing order.

Sample Input
2
6
5 4 4 4 5 4
8
6 5 6 5 6 5 5 6
Sample Output
3 6
2 4 8
 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <cmath>
 7 #include <time.h>
 8 #include <string>
 9 #include <map>
10 #include <stack>
11 #include <vector>
12 #include <set>
13 #include <queue>
14 #define inf 0x7fffffff
15 #define mod 10000
16 #define met(a,b) memset(a,b,sizeof a)
17 typedef long long ll;
18 using namespace std;
19 const int N = 100;
20 const int M = 100000;
21 const int INF = 0x3f3f3f3f;
22 int x;
23 int n, a[100002], b[100002];
24 int main()
25 {
26     int  t;
27     scanf("%d", &t);
28     while(t--)
29     {
30         scanf("%d", &n);
31         memset(a, 0, sizeof(a));
32         memset(b, 0, sizeof(b));
33         for(int i = 0; i < n; i++)
34         {
35             scanf("%d", &b[i]);
36             a[b[i]]++;
37         }
38         int c[1000];
39         int ans = 0;
40         for(int i = 1; i*i <= n; i++)
41         {
42             if(i*i == n) c[++ans] = i;
43             else if(n%i == 0)
44             {
45                 c[++ans] = i;
46                 c[++ans] = n/i;
47             }
48 
49         }
50         sort(c+1, c+1+ans);
51         for(int i = 1; i <= ans; i++)
52         {
53             int flag = 1;
54             for(int j = 1; j <= n; j++)
55             {
56                 if(a[j] > 0 && a[j]%c[i] != 0)
57                 {
58                     flag = 0;
59                     break;
60                 }
61             }
62             if(!flag) c[i] = -1;
63         }
64         int first = 1;
65         for(int i = ans; i >= 1; i--)
66         {
67             if(c[i] > 0)
68             {
69                 if(first)
70                 {
71                     first = 0;
72                     printf("%d", n/c[i]);
73                 }
74                 else printf(" %d", n/c[i]);
75             }
76 
77         }
78         printf("
");
79 
80     }
81 }

C、D待补
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原文地址：https://www.cnblogs.com/cshg/p/5926462.html