Codeforces Round · mfcheer

http://codeforces.com/contest/495

A. Digital Counter

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Malek lives in an apartment block with 100 floors numbered from 0 to 99 . The apartment has an elevator with a digital counter showing the floor that the elevator is currently on. The elevator shows each digit of a number with 7 light sticks by turning them on or off. The picture below shows how the elevator shows each digit.

One day when Malek wanted to go from floor 88 to floor 0 using the elevator he noticed that the counter shows number 89 instead of 88 . Then when the elevator started moving the number on the counter changed to 87 . After a little thinking Malek came to the conclusion that there is only one explanation for this: One of the sticks of the counter was broken. Later that day Malek was thinking about the broken stick and suddenly he came up with the following problem.

Suppose the digital counter is showing number n . Malek calls an integer x ( 0 ≤ x ≤ 99 ) good if it's possible that the digital counter was supposed to show x but because of some(possibly none) broken sticks it's showing n instead. Malek wants to know number of good integers for a specific n . So you must write a program that calculates this number. Please note that the counter always shows two digits.

Input

The only line of input contains exactly two digits representing number n ( 0 ≤ n ≤ 99 ). Note that n may have a leading zero.

Output

In the only line of the output print the number of good integers.

Sample test(s)

input

output

input

output

input

output

Note

In the first sample the counter may be supposed to show 88 or 89 .

In the second sample the good integers are 00 , 08 , 80 and 88 .

In the third sample the good integers are 03, 08, 09, 33, 38, 39, 73, 78, 79, 83, 88, 89, 93, 98, 99 .

A 打表，开始数错了一个WA了

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<stdlib.h>
#include<ctype.h>
#include<algorithm>
#include<vector>
#include<string.h>
#include<queue>
#include<stack>
#include<set>
#include<sstream>
#include<time.h>
#include<utility> 
#include<malloc.h> 

using namespace std;

char p[5];

int q[12] = {2,7,2,3,3,4,2,5,1,2};

int main()
{
    while (cin >> p)
    {
        int t = int (p[0] - '0');
        int tt = int (p[1] - '0');

        int ans = q[t] * q[tt];

        cout << ans << endl;
    }
    return 0;
}

B. Modular Equations

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Last week, Hamed learned about a new type of equations in his math class called Modular Equations. Lets define i modulo j as the remainder of division of i by j and denote it by . A Modular Equation, as Hamed's teacher described, is an equation of the form in which a and b are two non-negative integers and x is a variable. We call a positive integer x for which a solution of our equation.

Hamed didn't pay much attention to the class since he was watching a movie. He only managed to understand the definitions of these equations.

Now he wants to write his math exercises but since he has no idea how to do that, he asked you for help. He has told you all he knows about Modular Equations and asked you to write a program which given two numbers a and b determines how many answers the Modular Equation has.

Input

In the only line of the input two space-separated integers a and b ( 0 ≤ a , b ≤ 10 9 ) are given.

Output

If there is an infinite number of answers to our equation, print "infinity" (without the quotes). Otherwise print the number of solutions of the Modular Equation .

Sample test(s)

input

21 5

output

input

9435152 272

output

input

10 10

output

infinity

Note

In the first sample the answers of the Modular Equation are 8 and 16 since

B 优化算法提高效率以前接触过类似的

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<stdlib.h>
#include<ctype.h>
#include<algorithm>
#include<vector>
#include<string.h>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<sstream>
#include<time.h>
#include<utility>
#include<malloc.h>

using namespace std;

int n, m;

int slove(int x ,int y)
{
	int ans = 0;
	for (int i = 1; i*i <= x; i++)
	{
		if (i * i == x)
		{
			if (i > y)
				ans++;
		}
		else if (x % i == 0)
		{
			if (i > y)
			{
				ans++;
			}
			if (x/i > y)
			{
				ans++;
			}
		}
	}
	return ans;
}

int main()
{
	while (cin >> n >> m)
	{
		int ans = 0;
		n -= m;
		if (n == 0)
		{
			cout << "infinity" << endl;
			continue;
		}
		else if (n < 0)
		{
			cout << 0 << endl;
			continue;
		}
		else 
			cout << slove (n,m) << endl;
	}
	return 0; 
}

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