Three Garlands~Educational Codeforces Round 35
2 seconds
256 megabytes
standard input
standard output
Mishka is decorating the Christmas tree. He has got three garlands, and all of them will be put on the tree. After that Mishka will switch these garlands on.
When a garland is switched on, it periodically changes its state — sometimes it is lit, sometimes not. Formally, if i-th garland is switched on during x-th second, then it is lit only during seconds x, x + ki,x + 2ki, x + 3ki and so on.
Mishka wants to switch on the garlands in such a way that during each second after switching the garlands on there would be at least one lit garland. Formally, Mishka wants to choose three integersx1, x2 and x3 (not necessarily distinct) so that he will switch on the first garland during x1-th second, the second one — during x2-th second, and the third one — during x3-th second, respectively, and during each second starting from max(x1, x2, x3) at least one garland will be lit.
Help Mishka by telling him if it is possible to do this!
The first line contains three integers k1, k2 and k3 (1 ≤ ki ≤ 1500) — time intervals of the garlands.
If Mishka can choose moments of time to switch on the garlands in such a way that each second after switching the garlands on at least one garland will be lit, print YES.
Otherwise, print NO.
2 2 3
YES
4 2 3
NO
In the first example Mishka can choose x1 = 1, x2 = 2, x3 = 1. The first garland will be lit during seconds 1, 3, 5, 7, ..., the second — 2, 4, 6, 8, ..., which already cover all the seconds after the 2-nd one. It doesn't even matter what x3 is chosen. Our choice will lead third to be lit during seconds1, 4, 7, 10, ..., though.
In the second example there is no way to choose such moments of time, there always be some seconds when no garland is lit.
这题想不出了,于是暴力一发,就A了
暴力出奇迹啊
仔细想,这题确实傻逼
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <stack>
#include <string>
#include <math.h>
#include <vector>
using namespace std;
const int maxn = 2e5 + ;
const int INF = 0x7fffffff; int main() {
int a[];
while(scanf("%d%d%d", &a[], &a[], &a[]) != EOF) {
sort(a, a + );
if (a[] == ) {
printf("YES\n");
continue;
}
if (a[] == && a[] == ) {
printf("YES\n");
continue;
}
if (a[] == && a[] == && a[] == ) {
printf("YES\n");
continue;
}
if (a[] == && a[] == && a[] == ) {
printf("YES\n");
continue;
}
printf("NO\n");
}
return ;
}
Three Garlands~Educational Codeforces Round 35的更多相关文章
- Educational Codeforces Round 35 (Rated for Div. 2)
Educational Codeforces Round 35 (Rated for Div. 2) https://codeforces.com/contest/911 A 模拟 #include& ...
- Educational Codeforces Round 35 A. Nearest Minimums【预处理】
[题目链接]: Educational Codeforces Round 35 (Rated for Div. 2) A. Nearest Minimums time limit per test 2 ...
- Educational Codeforces Round 35 B/C/D
B. Two Cakes 传送门:http://codeforces.com/contest/911/problem/B 本题是一个数学问题. 有a个Ⅰ类球,b个Ⅱ类球:有n个盒子.将球放入盒子中,要 ...
- Educational Codeforces Round 35 (Rated for Div. 2)A,B,C,D
A. Nearest Minimums time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- Educational Codeforces Round 35
Nearest Minimums 相同的数里最小的数里的最小距离 Solution Two Cakes Solution Three Garlands 瞎比试 Solution Inversion C ...
- Educational Codeforces Round 35 B. Two Cakes【枚举/给盘子个数,两份蛋糕块数,最少需要在每个盘子放几块蛋糕保证所有蛋糕块都装下】
B. Two Cakes time limit per test 1 second memory limit per test 256 megabytes input standard input o ...
- 【Educational Codeforces Round 35 D】Inversion Counting
[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 排列中交换任意两个数字. 排列的逆序对个数的奇偶性会发生变化. 翻转这个过程其实就是len/2对数字发生交换. 交换了偶数次的话,不 ...
- 【Educational Codeforces Round 35 C】Two Cakes
[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 直觉题. 感觉情况会很少. 毕竟间隔太大了.中间肯定有一些数字达不到. 有1肯定可以 2 2 x肯定可以 3 3 3也可以 2 4 ...
- 【Educational Codeforces Round 35 B】Two Cakes
[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 从小到大枚举x. 看看a/x+b/x是不是大于等于n 是的话. 看看是不是两种蛋糕都能凑一堆. 也即x的最大枚举量是min(a,b) ...
随机推荐
- docker启用镜像常用脚本
语法:docker run [OPTIONS] IMAGE [COMMAND] [ARG...] OPTIONS说明:-a stdin: 指定标准输入输出内容类型,可选 STDIN/STDOUT/ST ...
- 实现BX的内容加上123 并把和送到寄存器AX
① 一条指令 ] ②两条指令 MOV AX,BX Tips: LEA指令与MOV指令的区别: ① MOV指令是 数据 传送指令-------传送数据 LEA指令是 有效地址 传送指令 ...
- netty-socket.io点对点通讯和聊天室通讯
netty-socketio是基于netty的socket.io服务实现,可以无缝对接前端使用的socketio-client.js. 相对于javaee的原生websocket支持(@serverE ...
- nuxt generate静态化后回退问题
之前线上的项目是nuxt build后的项目发布在服务器上,pm2来管理node的进程,nuxt还是运行在node的环境里. 这个方案用了半年左右,访问速度什么的确实很快,pm2管理下的node在wi ...
- 关于IT术语---ip、uv、pv、tps、qps、rps
涉及到IT方面的几条术语,这里要好好说道说道: 只要和网站打交道,难免会经常听到一系列的转有名词 >>> 系统今日UV多少.PV多少.QPS多少之类的问题.这里就对这些常见的术语 ...
- PHP表单安全过滤和防注入 htmlspecialchars() 和test_input()
什么是 htmlspecialchars() 函数? htmlspecialchars() 函数把特殊字符转换为 HTML 实体.这意味着 < 和 > 之类的 HTML 字符会被替换为 & ...
- 数据分析处理库Pandas——对象操作
Series结构 索引 修改 旧数据赋值给新数据,旧数据不变. 对某一数值进行修改,可以选择保留修改前或修改后的数值. 替换索引 修改某一个索引 添加 在数据1后添加数据2,数据1不改变. 添加一个数 ...
- C语言函数篇(五)静态库和动态库的创建和使用
使用库函数是源码的一种保护??? <我猜的.> 库函数其实不是新鲜的东西,我们一直都在用,比如C库. 我们执行pringf() 这个函数的时候,就是调用C库的函数. 下面记录静态库和动态库 ...
- [Bzoj4289]PA2012 Tax(Dijkstra+技巧建图)
Description 给出一个N个点M条边的无向图,经过一个点的代价是进入和离开这个点的两条边的边权的较大值,求从起点1到点N的最小代价.起点的代价是离开起点的边的边权,终点的代价是进入终点的边的边 ...
- poj 3111 卖珠宝问题 最大化平均值
题意:有N件分别价值v重量w的珠宝,希望保留k件使得 s=v的和/w的和最大 思路:找到贡献最大的 设当前的s为mid(x) 那么贡献就是 v-w*x 排序 ,取前k个 bool operator&l ...