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

Solution

将原串复制一遍加到后面，做一遍扩展KMP 。

扫一遍判断一下就能算出各种数的个数。

但是有一些是重复的数不应算在内。

于是我们可以算出原串的循环节长度（用类似KMP的方法， n−nex[n] 即为所求）。

用答案除以循环次数即可。

时间复杂度 O(T∗N) 。

Code

#include<cstdio> #include<cstring> using namespace std; const int N=2e5+5; int nex[N]; char s[N]; inline int max(int x,int y) { return x>y?x:y; } int main() { int T; scanf("%d",&T); for(int t=1;t<=T;t++) { scanf("%s",s+1); int n=strlen(s+1),m=n<<1,num=1; bool pd=true; for(int i=n+1;i<=m;i++) s[i]=s[i-n]; nex[1]=m; int mx=0,id=0; int ans1=0,ans2=1,ans3=0; for(int i=2;i<=n;i++) { int j=nex[i-id+1]; if(i+j>mx) { j=max(mx-i+1,0); while(i+j<=m && s[i+j]==s[j+1]) j++; } nex[i]=j; if(pd && nex[i]>=n-i+1) { if(n%(i-1)==0) num=n/(i-1); pd=false; } if(nex[i]>=n) ans2++; else if(s[nex[i]+1]>s[i+nex[i]]) ans1++; else ans3++; if(i+j-1>mx) mx=i+j-1,id=i; } printf("Case %d: %d %d %d\n",t,ans1/num,ans2/num,ans3/num); } return 0; }