MUV LUV EXTRA

Time Limit: 2000/1500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 134    Accepted Submission(s): 49

Problem Description One day, Kagami Sumika is stuck in a math problem aiming at calculating the length of a line segment with given statements and constraints. Since Sumika has no idea about it, she takes out a ruler and starts to measure the length. Unfortunately, the answer is an infinite decimal and she only got the first some digits of the answer from the ruler.
Sumika guesses that the answer is a rational number, which means that there exists two integers p, q that the answer equals qp. In this situation, the answer can be expressed as an infinte repeated decimal. For example, 12 = 0.500 ... , 13 = 0.333 .... , 910= 0.8999 ... ,3635= 1.0285714285714 ... .Sumika wants to guess the original number from the digits she got. Note that a number may has more than one way to be expressed such as 1.000 ... = 0.999 ... . Sumika won’t transform the digits she got to another form when guessing the original number.
Furthermore, Sumika relizes that for a repeating part, either too long or the appeared length too short will make the result unreliable. For example, if the decimal she measured is 1.0285714285714, it is obviously unreliable that the repeating part is “0285714285714”, since it is too long, or “428571”, since the appeared length is too short, which equals 7, the length of “4285714”. In this case, the best guess is “285714”, whose length is 6 and the appeared length is 12. So formally, she defines the reliability value of a repeating part, whose length is l and the appeared length is p, as the following formula:
a * p - b * l
Where a and b are given parameters.
Last but not least, you can ignore the integer parts of the decimal. It is just for restoring the scene. And the repeating part you guess should be completely repeated at least once and is still repeating at the end currently.
Please help Sumika determine the maximum reliability value among all repeating parts.  

 

Input The first line contains two positive integers a, b (1 ≤ a, b ≤ 109), denoting the parameters.
The next line contains a string s (1 ≤ |s| ≤ 107) in decimal form, denoting the first some digits of the accurate result.
It is guaranteed that there is exactly one decimal point in s and s is a legal non-negative decimal without leading "-"(the minus sign).  

 

Output Output a single line containing an integer, denoting the maximum reliability value.  

 

Sample Input 5 3 1.1020 2 1 12.1212  

 

Sample Output 9 6  

 

Source 642ccpcQHD  

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题解：

是个套了外壳的字符串题，求的是后缀的循环节长度和循环节出现的总长度(包含不完整循环节)，

显然反过来后跑一下kmp求出next数组，枚举前缀，那么i就是循环节出现的总长度，而i−next[i]就是循环节长度了，

直接更新最大值，答案可能有负数，初值要设为-1e18，2A。

参考代码：

#include<bits/stdc++.h>
#define maxl 10000010
using namespace std;

const long long inf=1ll<<61;
long long a,b,ans;
int slen,tlen;
int nxt[maxl];
char s[maxl],t[maxl];

inline void mainwork()
{
      ans=a-b;
      int j=0;nxt[1]=0;
      for(int i=2;i<=tlen;i++)
    {
          while((j && t[j+1]!=t[i]) || (j==tlen))
        j=nxt[j];
          if(t[j+1]==t[i] && j+1<=tlen)
        j++;
          nxt[i]=j;ans=max(ans,a*i-b*(i-nxt[i]));
    }
}

int main()
{
      while(~scanf("%lld%lld",&a,&b))
    {
          scanf("%s",s+1);
          slen=strlen(s+1);tlen=0;
          for(int i=slen;i>=1;i--)
            if(s[i]=='.')
                  break;
            else
                t[++tlen]=s[i];
          mainwork();
        printf("%lld\n",ans);
    }
      return 0;
}

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