专注收集记录技术开发学习笔记、技术难点、解决方案
网站信息搜索 >> 请输入关键词:
您当前的位置: 首页 > 编程

Codeforces Round #281 (Div. 二)E（数学）

发布时间：2011-06-30 11:49:30 文章来源：www.iduyao.cn 采编人员：星星草
Codeforces Round #281 (Div. 2)E（数学）
E. Vasya and Polynomial
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Vasya is studying in the last class of school and soon he will take exams. He decided to study polynomials. Polynomial is a function P(x) = a0 + a1x1 + ... + anxn. Numbers ai are called coefficients of a polynomial, non-negative integer n is called a degree of a polynomial.

Vasya has made a bet with his friends that he can solve any problem with polynomials. They suggested him the problem: "Determine how many polynomials P(x) exist with integer non-negative coefficients so that , and , where  and b are given positive integers"?

Vasya does not like losing bets, but he has no idea how to solve this task, so please help him to solve the problem.

Input

The input contains three integer positive numbers  no greater than 1018.

Output

If there is an infinite number of such polynomials, then print "inf" without quotes, otherwise print the reminder of an answer modulo109 + 7.

Sample test(s)
input
2 2 2
output
2
input
2 3 3
output
1

题意：RT

思路：给出了t,a,b,那么有

            f(t)=a0+a1*t+a2*t^2+...+an*t^n=a

           f(a)=a0+a1*a+a2*a^2+...+an*a^n=b

           a1+a2*t+...+an*t^(n-1)=(a-a0)/t

           a1+a2*a+...+an*a^(n-1)=(b-a0)/a

           那么(a-a0)%t=0 && (b-a0)%a=0

           a%t=a0%t && b%a=a0%a

           因为b>a

           所以b=k*a+a0;

          又因为a0<=a && a0<=b

          所以a0=b%a 或者 a0=b%a+a

          这样递归求解各个常数就可以了

友情提示：
信息收集于互联网，如果您发现错误或造成侵权，请及时通知本站更正或删除，具体联系方式见页面底部联系我们，谢谢。

其他相似内容：

热门推荐：