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 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.
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
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) ans ++; if((a-b)/x > b) ans ++; } } if((a-b) == x*x && x > b) ans ++; printf("%d\n", ans); } } return 0;}</time.h></stdlib.h></math.h></string></map></set></queue></vector></algorithm></iostream></string.h></stdio.h>
Python版：
a, b = map(int, raw_input().split())if a == b: print 'infinity'elif a b: ans += 1 if a/i > b and i*i != a: ans += 1 i += 1 print ans
