zoukankan      html  css  js  c++  java
  • B. Modular Equations

    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
    2
    
    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 

    这题的题意是求(n-m)比m大的因数个数,如果直接求会超时,所以可以用循环从1到i*i<=n-m;然后判断当前i和(n-m)/i是否大于m.

    #include<stdio.h>
    #include<string.h>
    int main()
    {
    	int n,m,i,j,num,t;
        while(scanf("%d%d",&n,&m)!=EOF)
    	{
    		if(n==m){
    			printf("infinity
    ");continue;
    		}
    		num=0;
    		for(i=1;i*i<=n-m;i++){
    			if((n-m)%i==0){
    				t=(n-m)/i;
    				if(t>m)num++;
    				if(i>m && i!=t)num++;
    			}
    		}
    		printf("%d
    ",num);
    	}
    }

  • 相关阅读:
    备忘录
    中缀表达式转为后缀表达式
    未出现的最小正整数
    摩尔投票算法
    两个等长升序序列找中位数
    Morris二叉树遍历
    2020牛客寒假算法基础集训营5 街机争霸
    2020牛客寒假算法基础集训营5 牛牛战队的比赛地
    2020牛客寒假算法基础集训营2 求函数
    2020牛客寒假算法基础集训营2 建通道
  • 原文地址:https://www.cnblogs.com/herumw/p/9464842.html
Copyright © 2011-2022 走看看