Codeforces Round #416(Div. 2)-811A.。。。 811B.。。。 811C.dp。。。不会
2 seconds
256 megabytes
standard input
standard output
At regular competition Vladik and Valera won a and b candies respectively. Vladik offered 1 his candy to Valera. After that Valera gave Vladik 2 his candies, so that no one thought that he was less generous. Vladik for same reason gave 3 candies to Valera in next turn.
More formally, the guys take turns giving each other one candy more than they received in the previous turn.
This continued until the moment when one of them couldn’t give the right amount of candy. Candies, which guys got from each other, they don’t consider as their own. You need to know, who is the first who can’t give the right amount of candy.
Single line of input data contains two space-separated integers a, b (1 ≤ a, b ≤ 109) — number of Vladik and Valera candies respectively.
Pring a single line "Vladik’’ in case, if Vladik first who can’t give right amount of candy, or "Valera’’ otherwise.
1 1
Valera
7 6
Vladik
Illustration for first test case:

Illustration for second test case:

题意就是两个人依次拿出多于上一个人的糖。
代码1:
#include<bits/stdc++.h>
using namespace std;
int main(){
int n,m;
int num1,num2;
while(~scanf("%d%d",&n,&m)){
num1=;num2=;
int t=n;
for(int i=;i<=t;i=i+){
if(n>=i+){
num1++;
n=n-(i+);
}
if(m>=i+){
num2++;
m=m-(+i);
}
}
//cout<<num1<<" "<<num2<<endl;
if(num1<=num2) printf("Vladik\n");
else printf("Valera\n");
}
return ;
}
代码2:
#include<bits/stdc++.h>
using namespace std;
int main(){
int n,m;
int num1,num2;
while(~scanf("%d%d",&n,&m)){
num1=;num2=;
int t=n;
for(int i=;i<=t;i++){ //我一开始写的i=0;i<=n,n会变的。
if(n>=*i-){
num1++;
n-=*i-;
}
if(m>=*i){
num2++;
m-=*i;
}
}
if(num1<=num2) printf("Vladik\n");
else printf("Valera\n");
}
return ;
}
Codeforces Round #416(Div. 2)-811A.。。。 811B.。。。 811C.dp。。。不会的更多相关文章
- Codeforces Round #174 (Div. 1) B. Cow Program(dp + 记忆化)
题目链接:http://codeforces.com/contest/283/problem/B 思路: dp[now][flag]表示现在在位置now,flag表示是接下来要做的步骤,然后根据题意记 ...
- Codeforces Round #416 (Div. 2)(A,思维题,暴力,B,思维题,暴力)
A. Vladik and Courtesy time limit per test:2 seconds memory limit per test:256 megabytes input:stand ...
- Codeforces Round#416 Div.2
A. Vladik and Courtesy 题面 At regular competition Vladik and Valera won a and b candies respectively. ...
- Codeforces Round #416 (Div. 2) C. Vladik and Memorable Trip
http://codeforces.com/contest/811/problem/C 题意: 给出一行序列,现在要选出一些区间来(不必全部选完),但是相同的数必须出现在同一个区间中,也就是说该数要么 ...
- Codeforces Round #416 (Div. 2) D. Vladik and Favorite Game
地址:http://codeforces.com/contest/811/problem/D 题目: D. Vladik and Favorite Game time limit per test 2 ...
- Codeforces Round #416 (Div. 2) A+B
A. Vladik and Courtesy 2 seconds 256 megabytes At regular competition Vladik and Valera won a and ...
- Codeforces Round #416 (Div. 2) B. Vladik and Complicated Book
B. Vladik and Complicated Book time limit per test 2 seconds memory limit per test 256 megabytes inp ...
- Codeforces Round #416 (Div. 2)A B C 水 暴力 dp
A. Vladik and Courtesy time limit per test 2 seconds memory limit per test 256 megabytes input stand ...
- 【分类讨论】【spfa】【BFS】Codeforces Round #416 (Div. 2) D. Vladik and Favorite Game
那个人第一步肯定要么能向下走,要么能向右走.于是一定可以判断出上下是否对调,或者左右是否对调. 然后他往这个方向再走一走就能发现一定可以再往旁边走,此时就可以判断出另一个方向是否对调. 都判断出来以后 ...
随机推荐
- Add to List 349. Intersection of Two Arrays
Given two arrays, write a function to compute their intersection. Example:Given nums1 = [1, 2, 2, 1] ...
- [知了堂学习笔记]_用JS制作《飞机大作战》游戏_第2讲(四大界面之间的跳转与玩家飞机的移动)
一.通过点击按钮事件,实现四大界面之间的跳转: (一)跳转的思路: 1.打开软件,只显示登录界面(隐藏游戏界面.暂停界面.玩家死亡界面) 2.点击微信登录(QQ登录)跳转到游戏界面,隐藏登录界面 3. ...
- ArcGIS API for Javascript 加载天地图(墨卡托投影)
<!DOCTYPE html> <html> <head> <meta charset="utf-8"> <title> ...
- 高仿二次元网易GACHA
高仿二次元网易GACHA,所有接口均通过Charles抓取而来,图片资源通过 https://github.com/yuedong56/Assets.carTool 工具提取. 详情见github地址 ...
- Swift3中数组创建方法
转载自:http://blog.csdn.net/bwf_erg/article/details/70858865 数组是由一组类型相同的元素构成的有序数据集合.数组中的集合元素是有 序的,而且可以重 ...
- MVC框架实例构建
转自:http://www.cnblogs.com/levenyes/p/3290885.html MVC全名是Model View Controller,是模型(model)-视图(view)-控制 ...
- Kotlin——最详细的操作符与操作符重载详解(上)
本篇文章为大家详细的介绍Koltin特有的操作符重载.或许对于有编程经验的朋友来说,操作符这个词绝对不陌生,就算没有任何编辑基础的朋友,数学中的算数运算符也绝不陌生.例如(+.-.*./.>.& ...
- python爬虫——写出最简单的网页爬虫
在我们日常上网浏览网页的时候,经常会看到一些好看的图片,我们就希望把这些图片保存下载,或者用户用来做桌面壁纸,或者用来做设计的素材.我们可以通过python 来实现这样一个简单的爬虫功能,把我们想要的 ...
- Nginx集群之WCF分布式消息队列
目录 1 大概思路... 1 2 Nginx集群之WCF分布式消息队列... 1 3 MSMQ消息队列... 2 4 编写WCF服务.客户端程序... ...
- C# log4net 的配置
项目的日志组件是必备可少的,任何项目中都需要.这样既方便前期的开发测试也方便项目后期的项目维护.C#项目的一个不错的日志组件是log4net,下面我就把桌面应用程序.控制台程序.网站中log4net的 ...