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.
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 ",ans); 46 return 0; 47 }