?? is practicing his program skill, and now he is given a string, he has to calculate the total number of its distinct substrings. 
But ?? thinks that is too easy, he wants to make this problem more interesting. 
?? likes a character X very much, so he wants to know the number of distinct substrings which contains at least one X. 
However, ?? is unable to solve it, please help him.

The first line of the input gives the number of test cases T;T test cases follow. 
Each test case is consist of 2 lines: 
First line is a character X, and second line is a string S. 
X is a lowercase letter, and S contains lowercase letters(‘a’-‘z’) only.

T<=30 
1<=|S|<=10^5 
The sum of |S| in all the test cases is no more than 700,000.

For each test case, output one line containing “Case #x: y”(without quotes), where x is the test case number(starting from 1) and y is the answer you get for that case.

FZU

2016 Multi-University Training Contest 4

wange2014   |   We have carefully selected several similar problems for you:  5808 5807 5806 5805 5804 

有一個子串X，一個母串str，求str中不相同的子串個數且至少含有一個X串。

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#define maxn 1000010

using namespace std;
//以下為倍增演算法求字尾陣列
int wa[maxn],wb[maxn],wv[maxn],Ws[maxn],nxt[maxn];
int sa[maxn],Rank[maxn],height[maxn];
char str[maxn],X[maxn];
int cmp(int *r,int a,int b,int l)
{
    return r[a]==r[b]&&r[a+l]==r[b+l];
}
void da(const char *r,int *sa,int n,int m)
{
    int i,j,p,*x=wa,*y=wb,*t;
    for(i=0; i<m; i++) Ws[i]=0;
    for(i=0; i<n; i++) Ws[x[i]=r[i]]++;
    for(i=1; i<m; i++) Ws[i]+=Ws[i-1];
    for(i=n-1; i>=0; i--) sa[--Ws[x[i]]]=i;
    for(j=1,p=1; p<n; j*=2,m=p)
    {
        for(p=0,i=n-j; i<n; i++) y[p++]=i;
        for(i=0; i<n; i++) if(sa[i]>=j) y[p++]=sa[i]-j;
        for(i=0; i<n; i++) wv[i]=x[y[i]];
        for(i=0; i<m; i++) Ws[i]=0;
        for(i=0; i<n; i++) Ws[wv[i]]++;
        for(i=1; i<m; i++) Ws[i]+=Ws[i-1];
        for(i=n-1; i>=0; i--) sa[--Ws[wv[i]]]=y[i];
        for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1; i<n; i++)
            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
    }
}

//求height陣列
void calheight(const char *r,int *sa,int n)
{
    int i,j,k=0;
    for(i=1; i<=n; i++) Rank[sa[i]]=i;
    for(i=0; i<n; height[Rank[i++]]=k)
        for(k?k--:0,j=sa[Rank[i]-1]; r[i+k]==r[j+k]; k++);
    return;
}

int slove(int n)
{
    int sum=0;
    for(int i=1; i<=n; i++)
        sum+=n-sa[i]-height[i];
    return sum;
}
int main()
{
    int t,ca=0;
    scanf("%d",&t);
    while(t--)
    {

        scanf("%s%s",X,str);
        int n=strlen(str);
        int temp=n;
        long long ans=0;
        for(int i=n-1; i>=0; i--)
        {
            if(str[i]==X[0])
            {
                temp=i; cout<<temp<<"===";
            }
           else cout<<temp<<"***";
            nxt[i]=temp;
        }
        da(str,sa,strlen(str)+1,130);
        calheight(str,sa,strlen(str));
        for(int i=1; i<=n; i++)
            // printf("%d %d\n",sa[i],height[i]);
            // printf("%d\n",slove(strlen(str)));
            ans+=n-max(nxt[sa[i]],sa[i]+height[i]);
        printf("Case #%d: %I64d\n",++ca,ans);
    }
    return 0;
}
/**
2
a
abc
b
bbb
**/