1353. Milliard Vasya's Function Time limit: 1.0 second Memory limit: 64 MB Vasya is the beginning mathematician. He decided to make an important contribution to the science and to become famous all over the world. But how can he do that if the most i…
1353. Milliard Vasya's Function Time limit: 1.0 second Memory limit: 64 MB Vasya is the beginning mathematician. He decided to make an important contribution to the science and to become famous all over the world. But how can he do that if the most i…
Vasya is studying number theory. He has denoted a function f(a, b) such that: f(a, 0) = 0; f(a, b) = 1 + f(a, b - gcd(a, b)), where gcd(a, b) is the greatest common divisor of a and b. Vasya has two numbers x and y, and he wants to calculate f(x, y).…
Discription Vasya is studying number theory. He has denoted a function f(a, b) such that: f(a, 0) = 0; f(a, b) = 1 + f(a, b - gcd(a, b)), where gcd(a, b) is the greatest common divisor of a and b. Vasya has two numbers x and y, and he wants to calcul…
数论题还是好恶心啊. 题目大意:给你两个不超过1e12的数 x,y,定义一个f ( x, y ) 如果y==0 返回 0 否则返回1+ f ( x , y - gcd( x , y ) ); 思路:我们设gcd ( x , y) 为G,那么 设 x = A*G,y = B*G,我们考虑减去多少个G时x y 的gcd会改变,我们设减去 k个G的时候 x和y 的gcd为改变,即 A*G 和 ( B - k ) * G 的 gcd 改变了,什么情况下会改变呢,就是A 和( B - k )的gcd…
Vasya follows a basketball game and marks the distances from which each team makes a throw. He knows that each successful throw has value of either 2 or 3 points. A throw is worth 2 points if the distance it was made from doesn't exceed some value of…
D - GCD of Polynomials 逆推,根据(i-2)次多项f(i-2)式和(i-1)次多项式f(i-1)推出i次多项式f(i) f(i)=f(i-1)*x+f(i-2) 样例已经给出0次和1次的了 注意系数绝对值大于1对2取模 代码: #include<bits/stdc++.h> using namespace std; #define ll long long #define pb push_back #define mem(a,b) memset(a,b,sizeof(a)…
Discription You are given a tree consisting of nn vertices. A number is written on each vertex; the number on vertex ii is equal to aiai. Let's denote the function g(x,y)g(x,y) as the greatest common divisor of the numbers written on the vertices bel…
A - Little Elephant and Function 思路: 水题: 代码: #include <cstdio> #include <iostream> using namespace std; int n; int main() { scanf("%d",&n); printf("%d ",n); ;i<n;i++) printf("%d ",i); ; }…
C. Enlarge GCD time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output Mr. F has n positive integers, a1,a2,-,an. He thinks the greatest common divisor of these integers is too small. So he wants to en…
