Codeforces Round #282 (Div. 2)-B. Modular Equations_html/css_WEB-ITnose

Modular Equations

time limit per test

memory limit per test

input

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  asolution 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?≤?109) are given.

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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

282

input

10 10

output

infinity

Note

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

GPT-MINUS1

通过在文本中随机地用同义词替换单词来愚弄GPT

83

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题意：给出a，b，问有多少满足a % x == b的正整数x存在。

分析：暴力可解。a % x == b有(a - b) % x == 0，也就是找a - b的因子。前提是：x是正整数，但是要注意需满足x > b（余数比除数小），当a

AC代码：

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;#define INF 0x7fffffffint main(){    #ifdef sxk        freopen("in.txt","r",stdin);    #endif    int a, b, ans;    while(scanf("%d%d",&a, &b)!=EOF)    {        ans = 0;        if(a < b) puts("0");        else if(a == b) puts("infinity");        else{            int x;            for(x=1; x*x<a-b; x++){                if((a-b) % x == 0){                    if(x > b) ans ++;                    if((a-b)/x > b) ans ++;                }            }            if((a-b) == x*x && x > b) ans ++;            printf("%d\n", ans);        }    }    return 0;}

Python版：

a, b = map(int, raw_input().split())if a == b:  print 'infinity'elif a < b:  print 0else:  a -= b  i = 1  ans = 0  while i*i <= a:    if a % i == 0:      if i > b:        ans += 1      if a/i > b and i*i != a:        ans += 1    i += 1  print ans