HDU 3374 String Problem （KMP+最大最小表示）

时间：2021-06-24 02:46:12

HDU 3374 String Problem （KMP+最大最小表示）

String Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1602    Accepted Submission(s): 714

Problem Description
Give you a string with length N, you can generate N strings by left shifts. For example let consider the string “SKYLONG”, we can generate seven strings:
String Rank 
SKYLONG 1
KYLONGS 2
YLONGSK 3
LONGSKY 4
ONGSKYL 5
NGSKYLO 6
GSKYLON 7
and lexicographically first of them is GSKYLON, lexicographically last is YLONGSK, both of them appear only once.
  Your task is easy, calculate the lexicographically fisrt string’s Rank (if there are multiple answers, choose the smallest one), its times, lexicographically last string’s Rank (if there are multiple answers, choose the smallest one), and its times also.
 
Input
  Each line contains one line the string S with length N (N <= 1000000) formed by lower case letters.
 
Output
Output four integers separated by one space, lexicographically fisrt string’s Rank (if there are multiple answers, choose the smallest one), the string’s times in the N generated strings, lexicographically last string’s Rank (if there are multiple answers, choose the smallest one), and its times also.
 
Sample Input
abcder
aaaaaa
ababab
 
Sample Output
1 1 6 1
1 6 1 6
1 3 2 3
 
 
这个题我开始直接做，感觉很容易，但WA
 
WA代码

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;

char str[1000005];
char str1[1000005];
int len;

int main(void)
{
int i;
int a1,a2;
while(scanf("%s",str)==1)
{
strcpy(str1,str);
len=strlen(str);
sort(str,str+len);
int k1=0,k2=0;
for(i=0;str[i]!='\0';i++)
{
if(str[i]==str[0])
k1++;
if(str[i]==str[len-1])
k2++;
}
for(i=0;str1[i]!='\0';i++)
{
if(str1[i]==str[0])
{
a1=i;
break;
}
}
for(i=0;str1[i]!='\0';i++)
{
if(str1[i]==str[len-1])
{
a2=i;
break;
}
}
printf("%d %d %d %d\n",a1+1,k1,a2+1,k2);
}
return 0;
}

不知道哪里错了，如果谁知道，请留言！

后来到网上搜了一下发现别人是用next数组，根据l=len-next[len]为最小字符串找个数，根据最小表示法，与最大表示法，找初始位置。

//最小表示法

int minexp(char *s,int x)//x=strlen(s);
{
int i=0,j=1,k=0,t;
while(i<x&&j<x&&k<x)
{
t=s[(i+k)%x]-s[(j+k)%x];
if(t==0) k++;
else
{
if(t>0)
i+=k+1;
else
j+=k+1;
if(i==j)
j++;
k=0;
}
}
return i<j?i:j;
}

 
 
//最大表示法
 

int maxexp(char *s,int x)
{
int i=0,j=1,k=0,t;
while(i<x&&j<x&&k<x)
{
t=s[(i+k)%x]-s[(j+k)%x];
if(t==0) k++;
else
{
if(t<0) i+=k+1; //这里是最小表示法与最大表示法的区别。
else j+=k+1;
if(i==j) j++;
k=0;
}
}
return i<j?i:j;
}

AC代码

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

char pat[1000005];
int lenp,next[1000005];

void get_next()
{
int i,j;
next[0]=-1;
i=0;
j=-1;
while(i<lenp)
{
if(j==-1||pat[i]==pat[j])
{
i++;
j++;
next[i]=j;
}
else
j=next[j];
}
}
//最小表示
int minexp(char *s,int x)
{
int i=0,j=1,k=0,t;
while(i<x&&j<x&&k<x)
{
t=s[(i+k)%x]-s[(j+k)%x];
if(t==0) k++;
else
{
if(t>0)
i+=k+1;
else
j+=k+1;
if(i==j)
j++;
k=0;
}
}
return i<j?i:j;
}

//最大表示
int maxexp(char *s,int x)
{
int i=0,j=1,k=0,t;
while(i<x&&j<x&&k<x)
{
t=s[(i+k)%x]-s[(j+k)%x];
if(t==0) k++;
else
{
if(t<0) i+=k+1; //这里是区别
else j+=k+1;
if(i==j) j++;
k=0;
}
}
return i<j?i:j;
}

int main()
{
int ans,i,j,a,b;
while(scanf("%s",pat)!=EOF)
{
lenp=strlen(pat);
get_next();
a=minexp(pat,lenp);
b=maxexp(pat,lenp);
int l=lenp-next[lenp];
if(lenp%l==0)
{
ans=lenp/l;//cout<<lenp/l<<endl;;
}
else
{
ans=1;//printf("1\n");
}
printf("%d %d %d %d\n",a+1,ans,b+1,ans);
}
return 0;
}

相关文章