A Simple Math Problem(HDU 1757 构造矩阵) If x < 10 f(x) = x.If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);And ai(0<=i<=9) can only be 0 or 1 . Sample Input...
HDU 5974 A Simple Math Problem Problem Description Given two positive integers a and b,find suitable X and Y to meet the conditions: X+Y=a Least Common Multiple (X, Y) =b Input Input includes multiple sets of test data.Each test da......
Lele now is thinking about a simple function f(x). If x < 10 f(x) = x. If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10); And ai(0<=i<=9) can only be 0 or 1 . Now, I will give a0 ~ a9 and two positive inte...
而已知x+y=ax+y=a 则可以推出x和y 代码 #include<bits/stdc++.h>#definefi first#definese second#definedebug cout<<"I AM HERE"<<endl;usingnamespacestd;typedeflonglongll;typedefpair<int,int> pii;constintmaxn=1e2+5,inf=0x3f3f3f3f,mod=1e9+7;constinteps=1e-3;inta,b;signedmain(){wh...
A Simple Math Problem 题目大意 输入描述 输出描述 样例 输入 3 输出 5 推导 #include<bits/stdc++.h> using namespace std; const int maxn = 1e6+1; int prime[maxn], mu[maxn], phi[maxn], tot; bool vis[maxn]; int f[maxn]; void get_pmp(int n) { vis[1] = mu[1] = phi...
HDU - 5974 A Simple Math Problem (数论解方程) Given two positive integers a and b,find suitable X and Y to meet the conditions: X+Y=a Least Common Multiple (X, Y) =b Input Input includes multiple sets of test data.Each test data occupies one line,including two positive integers a(...
HDU - 1757 A Simple Math Problem(基础矩阵快速幂) Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7158 Accepted Submission(s): 4373 题目描述: Problem Desc...HDU1757 A Simple Math Problem【矩阵快速幂】 A Simple Math Problem http:...
HDU1757:A Simple Math Problem(矩阵快速幂) Problem Description Lele now is thinking about a simple function f(x). If x < 10 f(x) = x. If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);...
#include<iostream> #include<stdio.h> #include<string.h> #include<math.h> #include<algorithm> #definepi 3.1415926 usingnamespacestd; intgcd(intm,intn) { if(n==0) returnm; else returngcd(n,m%n); } intmain() { inta,b; while(cin>>a>>b) ...
A Simple Math Problem Lele now is thinking about a simple function f(x). If x < 10 f(x) = x. If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10); And ai(0<=i<=9) can only be 0 or 1 . ...