Binary Witch
| Time Limit: 1000MS | Memory Limit: 65536K | |
Description
Once upon a time in the silent depths of digital forests there lived a Binary Witch. She was able to forecast weather, telling for any day in the future whether it will be rainy or sunny.
Her magic was based on the following ancient rule: let a1, a2, ..., aN be the sequence of binary digits, where ai = 0 indicates that i-th day was rainy, and ai = 1 -- that it was sunny. To predict the weather in day N+1, consider the t-postfix aN-t+1, aN-t+2, ..., aN consisting of the last t elements. If that postfix is encountered somewhere before the position N-t+1, i.e. if there is such k <= N-t, that ak = aN-t+1, ak+1 = aN-t+2, ..., ak+t-1 = aN then the predicted value will be ak+t.
If there is more than one occurrence of t-postfix, then the rightmost one (with maximal k) will be taken. So, to make a prediction, she tried t-postfixes, consequently for t = 13, 12, ..., 1, stopping after the first prediction. If neither postfix was found, she predicted rain ("0"). If prediction for more than one day is needed, it is assumed that all previous days are predicted correctly, so if first predicted value is b, then we make forecast for day N+2 based on N+1 values, where aN+1 = b.
Because the witch was burned long ago, your task is to write a program to perform her arcane job.
Her magic was based on the following ancient rule: let a1, a2, ..., aN be the sequence of binary digits, where ai = 0 indicates that i-th day was rainy, and ai = 1 -- that it was sunny. To predict the weather in day N+1, consider the t-postfix aN-t+1, aN-t+2, ..., aN consisting of the last t elements. If that postfix is encountered somewhere before the position N-t+1, i.e. if there is such k <= N-t, that ak = aN-t+1, ak+1 = aN-t+2, ..., ak+t-1 = aN then the predicted value will be ak+t.
If there is more than one occurrence of t-postfix, then the rightmost one (with maximal k) will be taken. So, to make a prediction, she tried t-postfixes, consequently for t = 13, 12, ..., 1, stopping after the first prediction. If neither postfix was found, she predicted rain ("0"). If prediction for more than one day is needed, it is assumed that all previous days are predicted correctly, so if first predicted value is b, then we make forecast for day N+2 based on N+1 values, where aN+1 = b.
Because the witch was burned long ago, your task is to write a program to perform her arcane job.
Input
First line of input file contains two integers N (1 <= N <= 1000000) and L (1 <= L <= 1000), separated by space. Second line contains a string of N characters "0" and "1".
Output
Output file must contain a single string of L characters, which are forecasts for days N+1, N+2, ..., N+L.
Sample Input
10 7 1101110010
Sample Output
0100100
Source
Northeastern Europe 2000, Far-Eastern Subregion
题意：
给一个长为n的字符串s，在前面找到一个长为m（m<=13）的字符串a=后缀
在m尽量大的前提下，使找到的字符串a尽量靠后
有则在字符串s末尾添加 字符串a的最后一个字符的后一个字符
没有则在字符串s末尾添加 0
连续找L次
最后输出字符串s的最后L位
逆序kmp
题目要求统计的是最长后缀，将原字符串翻转，就是最长前缀
然后对于接下来的每一天做一次kmp
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int maxn=1000000+1010; char s[maxn],str[maxn]; int next[maxn]; char ans[1005]; int n,l; int start=1005,len; void getnext(char *st) { int j,lm=strlen(st); for(int i=1;i<lm;i++) { j=next[i]; while(j&&st[j]!=st[i]) j=next[j]; next[i+1]= st[i]==st[j] ? j+1 : 0; } } int kmp(char *p) { int j=0,c=0,idx; int lm=strlen(p); for(int i=start+1;i<start+len;i++) { while(j&&p[j]!=str[i]) j=next[j]; if(p[j]==str[i]) j++; if(j>c) { c=j; idx=i; } if(j==lm) return i-j; } if(c) return idx-c; return -1; } int main() { char tmp[20]; scanf("%d%d",&n,&l); scanf("%s",s); len=strlen(s); for(int i=0;s[i];i++) str[start+i]=s[len-1-i]; int cnt=l,k; str[start+len]=' '; while(cnt--) { strncpy(tmp,str+start,13); if(len<13) tmp[len]=' '; else tmp[13]=' '; getnext(tmp); k=kmp(tmp); start--; len++; if(k==-1) str[start]='0'; else str[start]=str[k]; } for(int i=0;i<l;i++) ans[i]=str[start+l-1-i]; ans[l]=' '; printf("%s",ans); }