Codeforces Round #304 (Div. 2) Break the Chocolate 水题
Break the Chocolate
Time Limit: 20 Sec Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/546/problem/A
Description
A soldier wants to buy w bananas in the shop. He has to pay k dollars for the first banana, 2k dollars for the second one and so on (in other words, he has to pay i·k dollars for the i-th banana).
He has n dollars. How many dollars does he have to borrow from his friend soldier to buy w bananas?
Input
The first line contains three positive integers k, n, w (1 ≤ k, w ≤ 1000, 0 ≤ n ≤ 109), the cost of the first banana, initial number of dollars the soldier has and number of bananas he wants.
Output
Output one integer — the amount of dollars that the soldier must borrow from his friend. If he doesn't have to borrow money, output 0.
Sample Input
Sample Output
HINT
题意
你要买n个物品,物品的价格是i*k,然后问你还需要多少钱才行
题解:
啊,暴力暴力
代码:
//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
int Num;
char CH[];
//const int inf=0x7fffffff; //нчоч╢С
const int inf=0x3f3f3f3f;
/* inline void P(int x)
{
Num=0;if(!x){putchar('0');puts("");return;}
while(x>0)CH[++Num]=x%10,x/=10;
while(Num)putchar(CH[Num--]+48);
puts("");
}
*/
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
inline void P(int x)
{
Num=;if(!x){putchar('');puts("");return;}
while(x>)CH[++Num]=x%,x/=;
while(Num)putchar(CH[Num--]+);
puts("");
}
//************************************************************************************** int main()
{
ll k,n,w;
cin>>k>>n>>w;
ll ans=;
for(int i=;i<=w;i++)
ans+=k*i;
cout<<max(ans-n,0LL)<<endl;
}
Codeforces Round #304 (Div. 2) Break the Chocolate 水题的更多相关文章
- Codeforces Round #367 (Div. 2) A. Beru-taxi (水题)
Beru-taxi 题目链接: http://codeforces.com/contest/706/problem/A Description Vasiliy lives at point (a, b ...
- Codeforces Round #334 (Div. 2) A. Uncowed Forces 水题
A. Uncowed Forces Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/604/pro ...
- Codeforces Round #353 (Div. 2) A. Infinite Sequence 水题
A. Infinite Sequence 题目连接: http://www.codeforces.com/contest/675/problem/A Description Vasya likes e ...
- Codeforces Round #304 (Div. 2) A B C 水
A. Soldier and Bananas time limit per test 1 second memory limit per test 256 megabytes input standa ...
- Codeforces Round #306 (Div. 2) A. Two Substrings 水题
A. Two Substrings Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/550/pro ...
- Codeforces Round #327 (Div. 2) A. Wizards' Duel 水题
A. Wizards' Duel Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/591/prob ...
- Codeforces Round #360 (Div. 2) C. NP-Hard Problem 水题
C. NP-Hard Problem 题目连接: http://www.codeforces.com/contest/688/problem/C Description Recently, Pari ...
- Codeforces Round #146 (Div. 1) A. LCM Challenge 水题
A. LCM Challenge 题目连接: http://www.codeforces.com/contest/235/problem/A Description Some days ago, I ...
- Codeforces Round #335 (Div. 2) B. Testing Robots 水题
B. Testing Robots Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://www.codeforces.com/contest/606 ...
随机推荐
- ogg:Extract 进程遇长事务执行 Forcestop 引发的惨案
http://www.linuxidc.com/Linux/2015-04/115777.htm SQL> select t.addr,t.START_DATE from v$transacti ...
- Ubuntu之设置应用开机自启动
前言 前面使用oricle-Linux的时候,设置开机自启动使用的是chkconfig,现在使用ubuntu的时候发现Ubuntu系统没有了RH系统中的 chkconfig命令,因此研究了一下ubun ...
- 很多人都没用过的轻量级Oracle数据库数据导出工具SQLLDR2——性能超赞
SQLLDR2 介绍 每周发表一篇数据库或大数据相关的帖子,敬请关注 1. 工具介绍 Sqluldr2(SQL * UnLoader 第二版)是灵活与强大的 Oracle 文本导出程序,已被大众使 用 ...
- UOJ#58/BZOJ 3052【WC2013】糖果公园
好写好调的莫队算法,就算上树了仍然好写好调. 传送门 http://uoj.ac/problem/58 简要做法 将树按照dfs序分块,然后将询问按照(u所在块,v所在块,时间)作为关键字进行排序,依 ...
- 动态更新echarts k线图数据 通过websocket取数据
1.加载插件,实例化chart.2.链接websocket3.接收数据,处理数据,调用chart的实例,不断更新数据<!DOCTYPE html><html><head ...
- leetcode 之Longest Consecutive Sequence(六)
这题要仔细体会下哈希表的用法,要注意的是数组本身是无序的,因此需要向左右进行扩张. 另外这个思路可以进行聚类,把连续的标记为一类. int longestConsecutive(const vecto ...
- fiddler添加监测请求的 ip地址
本文转载自:http://www.jackness.org/2014/12/26/%E7%BB%99fiddler%E6%B7%BB%E5%8A%A0%E7%9B%91%E6%B5%8B%E8%AF% ...
- centos使用boost过程
1. 安装gcc,g++,make等开发环境 yum groupinstall "Development Tools" 2. 安装boost yum install boost ...
- touch命令的用法
touchtouch 文件,如果文件不存在,则创建一个新文件:如果文件存在,则将该存在的文件的修改时间或创建时间改为当前时间touch -t 时间戳 文件,则把该文件的时间改了
- 洛谷P2047 [NOI2007]社交网络 [图论,最短路计数]
题目传送门 社交网络 题目描述 在社交网络(social network)的研究中,我们常常使用图论概念去解释一些社会现象.不妨看这样的一个问题.在一个社交圈子里有n个人,人与人之间有不同程度的关系. ...