E. Jzzhu and Apples time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has picked n apples from his big apple tree. All the apples are numbered from 1 to n. Now he wants to sell them to…

[链接]:CF [题意]:对于一个数n,每次加一的代价是a,每次减一的代价是b,求被m整除时的最小代价. [分析]:分情况讨论,自己多举几个栗子. [代码]: #include<cstdio> #include<string> #include<cstdlib> #include<cmath> #include<iostream> #include<cstring> #include<set> #include<qu…

E. Jzzhu and Apples time limit per test: 1 seconds memory limit per test: 256 megabytes input: standard input output: standard output Jzzhu has picked \(n\) apples from his big apple tree. All the apples are numbered from \(1\) to \(n\). Now he wants…

E. Jzzhu and Apples time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has picked n apples from his big apple tree. All the apples are numbered from 1 to n. Now he wants to sell them to…

/* http://codeforces.com/problemset/problem/449/C cf 449 C. Jzzhu and Apples 数论+素数+贪心 */ #include <cstdio> #include <algorithm> using namespace std; ; int is_prime[Nmax]; int book[Nmax]; int cnt[Nmax]; int n,ans; void get__prime() { ;i<=n;i…

Codeforces Round #257 (Div. 1) C Codeforces Round #257 (Div. 1) E CF450E C. Jzzhu and Apples time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has picked n apples from his big apple tre…

Jzzhu and Apples 从大的质因子开始贪心, 如果有偶数个则直接组合, 如果是奇数个留下那个质数的两倍, 其余两两组合. #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair&l…

C. Jzzhu and Apples time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has picked n apples from his big apple tree. All the apples are numbered from 1 to n. Now he wants to sell them to…

cf 443 D. Teams Formation(细节模拟题) 题意: 给出一个长为\(n\)的序列,重复\(m\)次形成一个新的序列,动态消除所有k个连续相同的数字,问最后会剩下多少个数(题目保证消除的顺序对答案不会造成影响) \(n <= 10^{5} ,m <= 10^{9} ,k <= 10^{9},a_i <= 10^{5}\) 思路: 直接上题解好了 首先要用栈预处理序列本身,然后考虑处理后的序列 比较头和尾是否能够消除,细节需要处理好. #include<bi…

大意: 求从[1,n]范围选择尽量多的数对, 使得每对数的gcd>1 考虑所有除2以外且不超过n/2的素数p, 若p倍数可以选择的有偶数个, 直接全部划分即可 有奇数个的话, 余下一个2*p不划分, 其余全部划分 最后再将2的倍数全部划分一下即可 #include <iostream> #include <math.h> #include <string.h> #include <algorithm> #include <cstdio> #…