100197G Robbers
题目大意
看式子懂题意系列...
分析
自然想到我们先按比例下取整得到一个值,再按每个人这样分配所产生的值从大到小排序,然后将剩下的几个金币自大到小每人分配一个,代码挺好理解的,详见代码。
代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<ctime>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
#define li long long
#define pb push_back
#define mp make_pair
#define y1 y12345678909
#define rii register int
#define pii pair<int,int>
#define r(x) scanf("%d",&x)
#define ck(x) cout<<x<<endl;
#define uli unsigned long long
#define sp cout<<"---------------------------------------------------"<<endl
struct node {
int d,no;
}a[];
int ans[];
inline bool cmp (const node x,const node y){
return x.d<y.d;
}
int main(){
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
int n,m,i,z,y,all,x;
r(n),r(m),r(y);
all=m;
for(i=;i<=n;i++){
r(x);
z=x*m;
ans[i]+=z/y;
all-=ans[i];
z%=y;
a[i].d=z;
a[i].no=i;
}
sort(a+,a+n+,cmp);
for(i=n-all+;i<=n;i++)
ans[a[i].no]++;
for(i=;i<=n;i++)
printf("%d ",ans[i]);
puts("");
return ;
}
100197G Robbers的更多相关文章
- Codeforces Round #359 (Div. 2)C - Robbers' watch
C. Robbers' watch time limit per test 2 seconds memory limit per test 256 megabytes input standard i ...
- codeforces 359 C - Robbers' watch
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u Description Robb ...
- UVA1616-Caravan Robbers(枚举)
Problem UVA1616-Caravan Robbers Accept: 160 Submit: 1156Time Limit: 3000 mSec Problem Description O ...
- UVA1616-Caravan Robbers(二分)
Problem UVA1616-Caravan Robbers Accept: 96 Submit: 946Time Limit: 3000 mSec Problem Description Lon ...
- Codeforces 658A. Robbers' watch 模拟
A. Robbers' watch time limit per test: 2 seconds memory limit per test: 256 megabytes input: standar ...
- Codeforces Round #359 (Div. 1) A. Robbers' watch 暴力
A. Robbers' watch 题目连接: http://www.codeforces.com/contest/685/problem/A Description Robbers, who att ...
- SCOJ 4484 The Graver Robbers' Chronicles 后缀自动机
4484: The Graver Robbers' Chronicles 题目连接: http://acm.scu.edu.cn/soj/problem.action?id=4484 Descript ...
- Codeforces Round #359 (Div. 2) C. Robbers' watch 鸽巢+stl
C. Robbers' watch time limit per test 2 seconds memory limit per test 256 megabytes input standard i ...
- codeforces 686C C. Robbers' watch(dfs)
题目链接: C. Robbers' watch time limit per test 2 seconds memory limit per test 256 megabytes input stan ...
随机推荐
- github提交代码时遇到”Everything up-to-date“问题的解决方式
需要创建一个新分支,将最新代码加入新分支, 再将新分支合并到主分支,然后提交主分支代码到github网站. ---------------------------------------------- ...
- 高可用-软件heartbeat的入门介绍
注:参考互联网整理. 一.简介Linux-HA的全称是High-Availability Linux,它是一个开源项目,这个开源项目的目标是:通过社区开发者的共同努力,提供一个增强linux可靠性(r ...
- wordpress 插件 汉化
http://blog.wpjam.com/article/localizing-a-wordpress-plugin-using-poedit/ 翻译或者说本地化 WordPress 插件和主题可以 ...
- UVA 10417 Gift Exchanging
#include <iostream> #include <cstring> #include <stdio.h> #include <math.h> ...
- loj #6247. 九个太阳
求 $\sum\limits_{i=1}^n [k | i] \times C_n^i$ 膜 $998244353$ $n \leq 10^{15},k \leq 2^{20}$ $k$ 是 $2$ ...
- Mycat 在vscode中的开发配置
mycat是国产目前最被追捧的一款分布式数据库集群软件,有一些公司对数据库和应用都有自己的集群方案,但是更多的是一些面对庞大的数据量,而束手无策. 对于这种问题,我想百分之80遇到的是数据库的瓶颈,所 ...
- New Year and Counting Cards
Your friend has n cards. You know that each card has a lowercase English letter on one side and a di ...
- H国的身份证号码(搜索)
个人心得:巧妙利用数字进行维护就好了,深搜还是有点心得的: #1558 : H国的身份证号码I 时间限制:10000ms 单点时限:1000ms 内存限制:256MB 描述 H国的身份证号码是一个N位 ...
- LeetCode Base 7
原题链接在这里:https://leetcode.com/problems/base-7/#/description 题目: Given an integer, return its base 7 s ...
- Python函数-enumerate()
enumerate(sequence, [start=0]) 作用: 将可循环序列sequence以start开始分别列出序列数据和数据下标,即对一个可遍历的数据对象(如列表.元组或字符串),enum ...