首頁 > web前端 > html教學 > Codeforces Round #281 (Div. 2)E(数学)_html/css_WEB-ITnose

Codeforces Round #281 (Div. 2)E(数学)_html/css_WEB-ITnose

WBOY
發布: 2016-06-24 11:52:32
原創
1155 人瀏覽過

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
登入後複製

output

input

2 3 3
登入後複製

output


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


          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


來源:php.cn
本網站聲明
本文內容由網友自願投稿,版權歸原作者所有。本站不承擔相應的法律責任。如發現涉嫌抄襲或侵權的內容,請聯絡admin@php.cn
熱門教學
更多>
最新下載
更多>
網站特效
網站源碼
網站素材
前端模板