标签：ogr word order region 键值 == 字符 bar stand
Byteasar once decided to start manufacturing necklaces.
He subsequently bought a very long string of colourful coral beads for a bargain price.
Byteasar now also has a machine that, for a given 
(
), can cut the string into pieces (or substrings) of 
coral beads (i.e., the first piece consists of the beads no.
, the second of 
, etc.).
If the length of the string (measured in coral beads) is not a multiple of 
, then the last piece is not used, as it has length smaller than 
.
From now on we denote the colours of the beads with positive integers.
Byteasar, always praising diversity, wonders how he should choose the number 
in order to get as many different substrings as possible.
The ends of the long string that will be cut are different: there are specific beginning and ending (rather than two interchangeable endpoints), and the machine of course starts cutting at the beginning. On the other hand, in the substrings obtained from cutting the endpoints are interchangeable, and so the substrings can be reversed. In other words, the substrings 
and 
are identical to us. Write a program that determines the optimum value of 
for Byteasar.
Zxl有一次决定制造一条项链，她以非常便宜的价格买了一长条鲜艳的珊瑚珠子，她现在也有一个机器，能把这条珠子切成很多块（子串），每块有k（k>0）个珠子，如果这条珠子的长度不是k的倍数，最后一块小于k的就不要拉（nc真浪费），保证珠子的长度为正整数。 Zxl喜欢多样的项链，为她应该怎样选择数字k来尽可能得到更多的不同的子串感到好奇，子串都是可以反转的，换句话说，子串（1，2，3）和（3,2,1）是一样的。写一个程序，为Zxl决定最适合的k从而获得最多不同的子串。 例如：这一串珠子是： (1,1,1,2,2,2,3,3,3,1,2,3,3,1,2,2,1,3,3,2,1)， k=1的时候，我们得到3个不同的子串： (1),(2),(3) k=2的时候，我们得到6个不同的子串： (1,1),(1,2),(2,2),(3,3),(3,1),(2,3) k=3的时候，我们得到5个不同的子串： (1,1,1),(2,2,2),(3,3,3),(1,2,3),(3,1,2) k=4的时候，我们得到5个不同的子串： (1,1,1,2),(2,2,3,3),(3,1,2,3),(3,1,2,2),(1,3,3,2)
In the first line of the standard input there is an integer 
(
) denoting the length of the string to cut.
In the second line there are 
positive integers 
(
), separated by single spaces, that denote the colours of successive beads in Byteasar‘s string.
Two integers, separated by a single space, should be printed out to the first line of the standard ouput:
the (maximum) number of different substrings that can be obtained with an optimal choice of parameter 
, and the number 
of such optimal values of 
.
The second line should contain 
integers separated by single spaces:
the values of parameter 
that yield an optimum solution; these can be given in arbitrary order.
21
1 1 1 2 2 2 3 3 3 1 2 3 3 1 2 2 1 3 3 2 1
6 1
2
Solution：
本题字符串hash。
坑点在于子串可以反转，所以正反分别两次对原串hash，然后直接暴力枚举长度，问题出现在判重，想到用正反的hash值乘积做为新的键值，开一个set去重就好了。
代码：
1 #include<bits/stdc++.h> 2 #define il inline 3 #define ll long long 4 #define For(i,a,b) for(int (i)=(a);(i)<=(b);(i)++) 5 #define Bor(i,a,b) for(int (i)=(b);(i)>=(a);(i)--) 6 using namespace std; 7 const int N=200005,P=131,mod=1e9+7; 8 ll Ha[N],ha[N],sum[N],q,p; 9 unsigned ll H[N]; 10 int n,a[N],tot,ans[N],cnt; 11 set<unsigned ll>vis; 12 13 il int gi(){ 14 int a=0;char x=getchar(); 15 while(x<‘0‘||x>‘9‘)x=getchar(); 16 while(x>=‘0‘&&x<=‘9‘)a=(a<<3)+(a<<1)+x-48,x=getchar(); 17 return a; 18 } 19 20 int main(){ 21 n=gi(),sum[0]=1; 22 For(i,1,n) a[i]=gi(),sum[i]=sum[i-1]*P%mod,Ha[i]=(Ha[i-1]*P%mod+a[i])%mod; 23 Bor(i,1,n) ha[i]=(ha[i+1]*P%mod+a[i])%mod; 24 For(k,1,n){ 25 for(int i=1;i+k-1<=n;i+=k) 26 q=(Ha[i+k-1]-Ha[i-1]*sum[k]%mod+mod)%mod, 27 p=(ha[i]-ha[i+k]*sum[k]%mod+mod)%mod, 28 vis.insert(p*q); 29 int c=vis.size(); 30 if(c>tot)tot=c,ans[cnt=1]=k; 31 else if(c==tot) ans[++cnt]=k; 32 vis.clear(); 33 } 34 printf("%d %d\n",tot,cnt); 35 For(i,1,cnt) printf("%d ",ans[i]); 36 return 0; 37 }
标签：ogr word order region 键值 == 字符 bar stand
原文地址：https://www.cnblogs.com/five20/p/9470448.html
