0689 ZOJ - 4128

We call a string as a 0689-string if this string only consists of digits ‘0’, ‘6’, ‘8’ and ‘9’. Given a 0689-string s s s of length n n n, one must do the following operation exactly once: select a non-empty substring of s s s and rotate it 180 degrees.

More formally, let s i s_i si be the i i i-th character in string s s s. After rotating the substring starting from s l s_l sl and ending at s r s_r sr 180 degrees ( 1 ≤ l ≤ r ≤ n 1 \le l \le r \le n 1≤l≤r≤n), string s s s will become string t t t of length n n n extracted from the following equation, where t i t_i ti indicates the i i i-th character in string t t t: t i = { s i if 1 ≤ i < l or r < i ≤ n ’0’ if l ≤ i ≤ r and s l + r − i = ’0’ ’6’ if l ≤ i ≤ r and s l + r − i = ’9’ ’8’ if l ≤ i ≤ r and s l + r − i = ’8’ ’9’ if l ≤ i ≤ r and s l + r − i = ’6’ t_i = \begin{cases} s_i & \text{if } 1 \le i < l \text{ or } r < i \le n \\ \text{'0'} & \text{if } l \le i \le r \text{ and } s_{l+r-i} = \text{'0'} \\ \text{'6'} & \text{if } l \le i \le r \text{ and } s_{l+r-i} = \text{'9'} \\ \text{'8'} & \text{if } l \le i \le r \text{ and } s_{l+r-i} = \text{'8'} \\ \text{'9'} & \text{if } l \le i \le r \text{ and } s_{l+r-i} = \text{'6'} \\ \end{cases} ti=⎩⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎧si’0’’6’’8’’9’if 1≤i<l or r<i≤nif l≤i≤r and sl+r−i=’0’if l≤i≤r and sl+r−i=’9’if l≤i≤r and sl+r−i=’8’if l≤i≤r and sl+r−i=’6’

What’s the number of different strings one can get after the operation?

Input
There are multiple test cases. The first line of the input contains an integer T T T, indicating the number of test cases. For each test case:

The first and only line contains a 0689-string s s s ( 1 ≤ ∣ s ∣ ≤ 1 0 6 1 \le |s| \le 10^6 1≤∣s∣≤106).

It’s guaranteed that the sum of ∣ s ∣ |s| ∣s∣ of all test cases will not exceed 1 0 7 10^7 107.

Output
For each test case output one line containing one integer, indicating the number of different strings one can get after applying the operation exactly once.

Sample Input

2
0689
08

Sample Output

8
2

Code:

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<set>
#include<queue>
#include<map>
#include<stack>
#include<vector>
using namespace std;
typedef long long ll;
const int maxn = 1e5+5;
const int inf=0x3f3f3f3f;
const double PI=acos(-1.0);
const int mod=1e9+7;
string x;
int main()
{ll T;scanf("%lld",&T);while(T--){cin>>x;ll l=x.size();ll sum=(1+l)*l/2+1;ll a=0,b=0,xx=0,yy=0;for(ll i=0; i<l; i++){if(x[i]=='0')a++;if(x[i]=='8')b++;if(x[i]=='6')xx++;if(x[i]=='9')yy++;}if(xx==l||yy==l){printf("%lld\n",sum-1);continue;}sum-=(1+a)*a/2;sum-=(1+b)*b/2;sum-=xx*yy;printf("%lld\n",sum);}return 0;
}

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