Mod problem FZU - 2108 (recursion + fast exponentiation)

Given one non-negative integer A and one positive integer B, it’s very easy for us to calculate A Mod B. Here A Mod B means the remainder of the answer after A is divided by B. For example, 7 Mod 5 = 2, 12 Mod 3 = 0, 0 Mod 3 = 0.

In this problem, we use the following rules to express A.

(1) One non-empty string that only contains {0,1,2,3,4,5,6,7,8,9} is valid.

For example, 123, 000213, 99213. (Leading zeros is OK in this problem)

(2) If w is valid, then [w]x if valid. Here x is one integer that 0<x<10.

For example, [012]2=012012, [35]3[7]1=3535357.

(3) If w and v are valid, then wv is valid.

For example, w=[231]2 and v=1, then wv=[231]21 is valid (which is 2312311).

Now you are given A and B. Here A is express as the rules above and B is simply one integer, you are expected to output the A Mod B.

Input

The first line of the input contains an integer T(T≤10), indicating the number of test cases.

Then T cases, for any case, only two lines.

The first line is one non-empty and valid string that expresses A, the length of the string is no more than 1,000.

The second line is one integer B(0<B<2,000,000,000).

You may assume that the length of number A in decimal notation will less than 2^63.

Output

For each test case, output A Mod B in a single line.

Sample Input

3
[0]9[[1]2]3
10007
[[213231414343214231]5]1
10007
[0012]1
1

Sample Output

1034
3943

0

The meaning of the question: Given a string, it represents a number, the representation method is [12]x, x is a single digit, which represents the number of repetitions, find the number for b and take the modulus b

#include <cstdio>
#include <cmath>
#include <iostream>
#include <cstring>
#include <string>

using namespace std;

typedef long long ll;

char s[1000006];
ll md;

ll qm(ll a,ll n)
{
    ll ans = 1;
	while(n)
	{
		if(n%2)
		{
		    years = (years*y) % md;	
		}	
		a = (a*a)%md;
		n /= 2;
	}	
	return ans;
}

void dfs(int l,int r,ll &k,ll &v)
{
	k = 0;
	v = 0;
	int b = 0;
	int x,y;
	for(int i = l; i <= r; i++)
	{
		if(s[i] == '[')
		{
			if(b == 0) x = i+1;
			b++;
		}
		else if(s[i] == ']')
		{
			b--;
			if(b == 0) y = i-1;
			else continue;
			i++;
			ll t = s[i]-'0';
			ll ck,cv;
			dfs(x,y,ck,cv);
			k += ck*t;
			ll e = qm(10,ck);
			while(t--) v = (v*e+cv)%md;
		}
		else if(!b)
		{
			v = (v*10+s[i]-'0')%md;
			k++;
		}
	}
}

intmain()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s%d",s,&md);
        ll k,v;
        int len ​​= strlen(s);
        dfs (0, len-1, k, v);
        cout << v <<endl;
	}
    return 0;	
}