Revolving Digits

Revolving Digits

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 26557    Accepted Submission(s): 5870

Problem Description

One day Silence is interested in revolving the digits of a positive integer. In the revolving operation, he can put several last digits to the front of the integer. Of course, he can put all the digits to the front, so he will get the integer itself. For example, he can change 123 into 312, 231 and 123. Now he wanted to know how many different integers he can get that is less than the original integer, how many different integers he can get that is equal to the original integer and how many different integers he can get that is greater than the original integer. We will ensure that the original integer is positive and it has no leading zeros, but if we get an integer with some leading zeros by revolving the digits, we will regard the new integer as it has no leading zeros. For example, if the original integer is 104, we can get 410, 41 and 104.

 

Input

The first line of the input contains an integer T (1<=T<=50) which means the number of test cases.
For each test cases, there is only one line that is the original integer N. we will ensure that N is an positive integer without leading zeros and N is less than 10^100000.

 

Output

For each test case, please output a line which is "Case X: L E G", X means the number of the test case. And L means the number of integers is less than N that we can get by revolving digits. E means the number of integers is equal to N. G means the number of integers is greater than N.

 

Sample Input

1 341

 

Sample Output

Case 1: 1 1 1

 

Source

2012 Multi-University Training Contest 4

 

題意：給你一個數字，把最後一個數字放到最前面去，經過幾次變換後又回到原數字，問在這些數字中，比原數字小的，相等的，大的分別有多少個，去除重複的

擴展kmp尋找移位前和移位後的字符串要比較的唯一確定的位置。  去除重複的數字用kmp找到循環節

代碼如下：

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
const int N=100500;
char s[2*N],t[N];
int ext[N],extand[N];
int slen,tlen,len;
int get_len(char t[])
{
    int i=0,j=-1;
    ext[0]=-1;
    while(i<tlen)
    {
        if(j==-1||t[i]==t[j])
            ext[++i]=++j;
        else
            j=ext[j];
    }
    return tlen-ext[tlen];
}
void getnext(char T[]){// next[i]: 以第i位置開始的子串 與 T的公共前綴
     int i,length = strlen(T);
     ext[0] = length;
     for(i = 0;i<length-1 && T[i]==T[i+1]; i++);
          ext[1] = i;
          int a = 1;
          for(int k = 2; k < length; k++){
                  int p = a+ext[a]-1, L = ext[k-a];
                  if( (k-1)+L >= p ){
                       int j = (p-k+1)>0? (p-k+1) : 0;
                       while(k+j<length && T[k+j]==T[j]) j++;// 枚舉(p+1，length) 與(p-k+1,length) 區間比較
                       ext[k] = j, a = k;
                  }
                  else ext[k] = L;
         }
}


void getextand(char S[],char T[]){
    memset(ext,0,sizeof(ext));
    getnext(T);
    int  a = 0;
    int MinLen = slen>tlen?tlen:slen;
    while(a<MinLen && S[a]==T[a]) a++;
    extand[0] = a, a = 0;
    for(int k = 1; k < slen; k++){
        int p = a+extand[a]-1,L= ext[k-a];
        if( (k-1)+L >= p )
        {
            int j = (p-k+1)>0? (p-k+1) : 0;
            while(k+j<slen && j<tlen && S[k+j]==T[j] ) j++;
            extand[k] = j;a = k;
        }
        else extand[k] = L;
    }
}

int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%s",t);
        strcpy(s,t);
        strcat(s,t);
        slen=strlen(s);
        tlen=strlen(t);
        len=get_len(t);//循環節
        if(tlen%len!=0)
            len=tlen;
        getextand(s,t);
        int s1(0),s2(0),s3(0);
        for(int j=0;j<len;j++)
        {
            if(extand[j]>=tlen)//也可以不用判斷 s2一定爲1
                s2++;
            else if(s[extand[j]+j]<s[extand[j]])
                s1++;
            else if(s[extand[j]+j]>s[extand[j]])
                s3++;
        }
        printf("Case %d: ",i);
        printf("%d %d %d\n",s1,s2,s3);
    }
    return 0;
}