HDU 5442 Favorite Donut

Problem Description

Lulu has a sweet tooth. Her favorite food is ring donut. Everyday she buys a ring donut from the same bakery. A ring donut is consists of 

n

 parts. Every part has its own sugariness that can be expressed by a letter from 

a

 to 

z

 (from low to high), and a ring donut can be expressed by a string whose i-th character represents the sugariness of the 

i−th

 part in clockwise order. Note that 

z

 is the sweetest, and two parts are equally sweet if they have the same sugariness.

Once Lulu eats a part of the donut, she must continue to eat its uneaten adjacent part until all parts are eaten. Therefore, she has to eat either clockwise or counter-clockwise after her first bite, and there are 

2n

 ways to eat the ring donut of 

n

 parts. For example, Lulu has 

6

 ways to eat a ring donut 

abc

: 

abc,bca,cab,acb,bac,cba

. Lulu likes eating the sweetest part first, so she actually prefer the way of the greatest lexicographic order. If there are two or more lexicographic maxima, then she will prefer the way whose starting part has the minimum index in clockwise order. If two ways start at the same part, then she will prefer eating the donut in clockwise order. Please compute the way to eat the donut she likes most.

Input

First line contain one integer 

T,T≤20

, which means the number of test case.

For each test case, the first line contains one integer 

n,n≤20000

, which represents how many parts the ring donut has. The next line contains a string consisted of 

n

 lowercase alphabets representing the ring donut.

Output

You should print one line for each test case, consisted of two integers, which represents the starting point (from 

1

 to 

n

) and the direction (

 for clockwise and 

1

 for counterclockwise).

Sample Input

2

4

abab

4

aaab

Sample Output

2 0

4 0

最小表示法

#include<cstdio>
#include<cstring>
#include<ctime>
#include<algorithm>
using namespace std;
const int maxn = 100001;
int n,T,f[maxn];
char s[maxn],S[maxn];

int cmp(int x, char a, char b)
{
    if (a > b) return x - 0;
    if (a < b) return 1 - x;
    return 2;
}

int present(int z)
{
    int i, j, k, x, y;
    for (i = 0, j = 1, k = 0; k < n;)
    {
        if (i == j) { j++; k = 0; }
        x = (i + k) % n;    y = (j + k) % n;
        switch (cmp(z, s[x], s[y]))
        {
            case 0:i += k + 1; k = 0; break;
            case 1:j += k + 1; k = 0; break;
            case 2:k++; break;
        }
    }
    return min(i, j);
}

int Present(int z,int a,int b)
{
    int i, j, k, x, y;
    //int tot=n;
    //memset(f,0,sizeof(f));
    for (i = a, j = b, k = 0; k < n;)
    {
        if (i == j) { j--; k = 0; }
        x = ((i - k) % n + n)%n;
        y = ((j - k) % n + n)%n;
        switch (cmp(z, s[x], s[y]))
        {
            case 0:i -= k + 1; k = 0; break;
            case 1:j -= k + 1; k = 0; break;
            case 2:k++; break;
        }
    }
    return max(i, j);
}

int get()
{
    int l=1,r=n-1,mid,ans;
    while (l<r)
    {
        mid=(l+r)>>1;
        ans=Present(1,mid,mid-1);
        if (ans>=0) r=mid;
        else l=mid+1;
    }
    ans=Present(1,r,r-1);
    return ans;
}

int main()
{
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d%s",&n,s);
        if (n==1) printf("1 0\n");
        else 
        {
            int v1=present(1);
            int v2=get();
            int flag=-1;
            for (int i=0;i<n;i++)
            {
                char c1=s[(v1+i)%n];
                char c2=s[(v2-i+n)%n];
                if (c1>c2) {flag=0;    break;}
                if (c1<c2) {flag=1; break;}
            }
            if (flag==-1)
            {
                if (v1>v2) flag=1;
                else flag=0;
            }
            if (flag) printf("%d %d\n",v2+1,flag);
            else printf("%d %d\n",v1+1,flag);
        }
    }
    return 0;
}