String Problem HDU - 3374 (最大最小表示法)

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原创最新推荐文章于 2026-03-01 03:44:01 发布 309 阅读 AI 写同款· GEO 优化 ›

Problem E: 求最大值和最小值 Description 求出一些整数中的最大值和最小值。 Input 输入为多行，以EOF结束。每行为一个十进制的数字，全部由0～9组成，每行最长不超过1000个字符。有些数字会以0开头（不影响数值大小的前导0）。 Output 输出为输入中的最大值和最小值，格式见sample。 Sample Input 02010001201223 Sample Ou 阅读详情

String Problem

HDU - 3374

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

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code：

#include <iostream>
#include <cstdio>
#include <cstring>
#define N 1000005
using namespace std;
char s[N];
int Next[N];
void getNext(){
    int i = -1,j = 0;
    memset(Next,0,sizeof(Next));
    Next[0] = -1;
    int len = strlen(s);
    while(j < len){
        if(i == -1 || s[i] == s[j]){
            i++,j++;
            Next[j] = i;
        }
        else{
            i = Next[i];
        }
    }
}
int min_max_express(bool flag){
    int i = 0,j = 1, k = 0;
    int len = strlen(s);
    while(i < len && j < len && k < len){
        int t = s[(j+k)%len]-s[(i+k)%len];
        if(t == 0)
            k++;
        else{
            if(flag){
                if(t > 0) j = j+k+1;
                else i = i+k+1;
            }
            else{
                if(t > 0) i = i+k+1;
                else j = j+k+1;
            }
            if(i == j) j++;
            k = 0;
        }
    }
    int ans = min(i,j);
    return ans;
}
int main(){
    while(~scanf("%s",s)){
        getNext();
        int mins = min_max_express(true);
        int maxs = min_max_express(false);
        int len = strlen(s);
        int tmp = len-Next[len];
        int times = len % tmp ? 1 : len/tmp;
        printf("%d %d %d %d\n",mins+1,times,maxs+1,times);
    }
    return 0;
}

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