标签：class 内存 ima += orm out string 字符串哈希 for
Isomorphic Inversion
时间限制: 1 Sec 内存限制: 128 MB
题目描述
Let s be a given string of up to 106 digits. Find the maximal k for which it is possible to partition s into k consecutive contiguous substrings, such that the k parts form a palindrome.
More precisely, we say that strings s0, s1, . . . , sk−1 form a palindrome if si = sk−1−i for all 0 ≤ i < k.
In the first sample case, we can split the string 652526 into 4 parts as 6|52|52|6, and these parts together form a palindrome. It turns out that it is impossible to split this input into more than 4 parts while still making sure the parts form a palindrome.
输入
A nonempty string of up to 106 digits.
输出
Print the maximal value of k on a single line.
样例输入
652526
样例输出
4
题意：给一串数字串，求问最多能分成多少个区域使得区域回文。
1 #include<bits/stdc++.h>
2 #pragma GCC optimize(3)
3 using namespace std;
4 typedef long long ll;
5 const int maxn=1e6+7;
6 const ll prime=97;
7 ll Ha[maxn];
8 char str[maxn];
9 void init()
10 {
11 Ha[0]=1;
12 for(int i=1; i<=maxn-5; ++i)
13 {
14 Ha[i]=Ha[i-1]*prime;
15 }
16 }
17 int main()
18 {
19 init();
20 int ans=0;
21 scanf("%s",str+1);
22 int len=strlen(str+1);
23 int l=1,r=len;
24 while(l<=r)
25 {
26 int start=r;
27 ll ldata=ll(str[l]-‘0‘),rdata=ll(str[r]-‘0‘);
28 while(ldata!=rdata&&(l<r))
29 {
30 ++l;
31 --r;
32 ldata=ll(str[l]-‘0‘)+ldata*prime;
33 rdata+=ll(str[r]-‘0‘)*Ha[start-r];
34 }
35 if(ldata!=rdata)
36 {
37 ++ans;
38 break;
39 }
40 if(l!=r)ans+=2;
41 else ++ans;
42 ++l;
43 --r;
44 }
45 printf("%d\n",ans);
46 return 0;
47 }
BAPC 2018 Preliminaries-Isomorphic Inversion（字符串哈希）
标签：class 内存 ima += orm out string 字符串哈希 for
原文地址：https://www.cnblogs.com/CharlieWade/p/11451753.html
