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

題意：給出多組數據，每組數據給出一個字符串，要求輸出這個字符串的最小最大表示的起始位置，然后分別求出在同構串中起始位置的字符出現的次數

思路：最小最大的起始位置直接套用模版即可，然后使用 KMP 的 next 數組求循環節，則次數=長度/循環節長度

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #define PI acos(-1.0) #define E 1e-9 #define INF 0x3f3f3f3f #define LL long long const int MOD=10007; const int N=1000000+5; const int dx[]= {-1,1,0,0}; const int dy[]= {0,0,-1,1}; using namespace std; int Next[N]; char str[N]; void getNext(char p[]){Next[0]=-1;int len=strlen(p);int j=0;int k=-1;while(j<len){if(k==-1||p[j]==p[k]){k++;j++;Next[j]=k;}else{k=Next[k];}} } int minmumRepresentation(char *str){//最小表示法int len=strlen(str);int i=0;int j=1;int k=0;while(i<len&&j<len&&k<len){int temp=str[(i+k)%len]-str[(j+k)%len];if(temp==0)k++;else{if(temp>0)i=i+k+1;elsej=j+k+1;if(i==j)j++;k=0;}}return i<j?i:j; } int maxmumRepresentation(char *str){//最大表示法int len=strlen(str);int i=0;int j=1;int k=0;while(i<len&&j<len&&k<len){int temp=str[(i+k)%len]-str[(j+k)%len];if(temp==0)k++;else{if(temp>0)j=j+k+1;elsei=i+k+1;if(i==j)j++;k=0;}}return i<j?i:j; }int main(){while(scanf("%s",str)!=EOF){getNext(str);int n=strlen(str);int len=n-Next[n];int num=1;//數量if(n%len==0)num=n/len;int minn=minmumRepresentation(str);//最小表示int maxx=maxmumRepresentation(str);//最大表示printf("%d %d %d %d\n",minn+1,num,maxx+1,num);}return 0; }

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