日韩性视频-久久久蜜桃-www中文字幕-在线中文字幕av-亚洲欧美一区二区三区四区-撸久久-香蕉视频一区-久久无码精品丰满人妻-国产高潮av-激情福利社-日韩av网址大全-国产精品久久999-日本五十路在线-性欧美在线-久久99精品波多结衣一区-男女午夜免费视频-黑人极品ⅴideos精品欧美棵-人人妻人人澡人人爽精品欧美一区-日韩一区在线看-欧美a级在线免费观看

歡迎訪問 生活随笔!

生活随笔

當(dāng)前位置: 首頁 > 编程资源 > 编程问答 >内容正文

编程问答

39行代码AC_HDU-6740 2019CCPC秦皇岛 J MUV LUV EXTRA(KMP变形)

發(fā)布時間:2024/2/28 编程问答 39 豆豆
生活随笔 收集整理的這篇文章主要介紹了 39行代码AC_HDU-6740 2019CCPC秦皇岛 J MUV LUV EXTRA(KMP变形) 小編覺得挺不錯的,現(xiàn)在分享給大家,幫大家做個參考.

勵志用少的代碼做高效表達(dá)


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?la * p - b * la?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.


解題思路

如果有對KMP還不懂的同學(xué)——>KMP算法_圖示分析+解析+例題

這個題很巧妙的利用了 KMP

我們需要反過來求kmp得next數(shù)組,枚舉前綴,那么i就是循環(huán)節(jié)出現(xiàn)的總長度,而i-next[i]就是循環(huán)節(jié)長度。


代碼展示

#include<bits/stdc++.h> using namespace std; typedef long long ll; const ll INF=1e18; ll a,b,res; int i,j,k; char s[10000010], s1[10000010]; int Next[10000010];void getnext(char s[],int Next[]) {int q,k;int len=strlen(s);Next[0] = 0;for (q = 1,k = 0; q < len; ++q) {while(k>0 && s[q]!=s[k]) k = Next[k-1];if (s[q] == s[k]) k++;Next[q] = k;} }int main() {while(cin>>a>>b) {memset(Next,0,sizeof Next);cin>>s1;int len1=strlen(s1);int len=0;for(i=len1-1; i>=0; i--) {if(s1[i]=='.') break;s[len1-i-1]=s1[i];len++;}getnext(s,Next);ll ans=-INF;for(i=0; i<len; i++) {res=a*(i+1)-b*(i+1-Next[i]); //核心代碼if(res>ans) ans=res;}cout << ans << '\n';} return 0; }

總結(jié)

以上是生活随笔為你收集整理的39行代码AC_HDU-6740 2019CCPC秦皇岛 J MUV LUV EXTRA(KMP变形)的全部內(nèi)容,希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯,歡迎將生活随笔推薦給好友。