A. GukiZ and Contest
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Professor GukiZ likes programming contests. He especially likes to rate his students on the contests he prepares. Now, he has decided to prepare a new contest.

In total, n students will attend, and before the start, every one of them has some positive integer rating. Students are indexed from 1 to n. Let's denote the rating of i-th student as ai. After the contest ends, every student will end up with some positive integer position. GukiZ expects that his students will take places according to their ratings.

He thinks that each student will take place equal to . In particular, if student A has rating strictly lower then student BA will get the strictly better position than B, and if two students have equal ratings, they will share the same position.

GukiZ would like you to reconstruct the results by following his expectations. Help him and determine the position after the end of the contest for each of his students if everything goes as expected.

Input

The first line contains integer n (1 ≤ n ≤ 2000), number of GukiZ's students.

The second line contains n numbers a1, a2, ... an (1 ≤ ai ≤ 2000) where ai is the rating of i-th student (1 ≤ i ≤ n).

Output

In a single line, print the position after the end of the contest for each of n students in the same order as they appear in the input.

Examples
input
3
1 3 3
output
3 1 1
input
1
1
output
1
input
5
3 5 3 4 5
output
4 1 4 3 1
Note

In the first sample, students 2 and 3 are positioned first (there is no other student with higher rating), and student 1 is positioned third since there are two students with higher rating.

In the second sample, first student is the only one on the contest.

In the third sample, students 2 and 5 share the first position with highest rating, student 4 is next with third position, and students 1 and 3are the last sharing fourth position.

问你一个数排第几

#include<bits/stdc++.h>
using namespace std;
int a[];
int main(){
int n;
cin>>n;
for(int i=;i<n;i++)
cin>>a[i];
int s=;
for(int j=;j<n;j++)
if(a[j]>a[])s++;
printf("%d",s);
for(int i=;i<n;i++){
s=;
for(int j=;j<n;j++)
if(a[j]>a[i])s++;
printf(" %d",s);}
return ;
}

直接这样枚举也能过,所以,sort的也能过吧

#include<bits/stdc++.h>
using namespace std;
int a[],b[];
int main(){
int n;
cin>>n;
for(int i=;i<n;i++)
cin>>a[i];
copy(a,a+n,b);
sort(b,b+n);
cout<<n-(upper_bound(b,b+n,a[])-b)+;
for(int i=;i<n;i++)
cout<<" "<<n-(upper_bound(b,b+n,a[i])-b)+;
return ;
}

完美,直接upper_bound美滋滋

B. ZgukistringZ
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Professor GukiZ doesn't accept string as they are. He likes to swap some letters in string to obtain a new one.

GukiZ has strings ab, and c. He wants to obtain string k by swapping some letters in a, so that k should contain as many non-overlapping substrings equal either to b or c as possible. Substring of string x is a string formed by consecutive segment of characters from x. Two substrings of string x overlap if there is position i in string x occupied by both of them.

GukiZ was disappointed because none of his students managed to solve the problem. Can you help them and find one of possible strings k?

Input

The first line contains string a, the second line contains string b, and the third line contains string c (1 ≤ |a|, |b|, |c| ≤ 105, where |s| denotes the length of string s).

All three strings consist only of lowercase English letters.

It is possible that b and c coincide.

Output

Find one of possible strings k, as described in the problem statement. If there are multiple possible answers, print any of them.

Examples
input
aaa
a
b
output
aaa
input
pozdravstaklenidodiri
niste
dobri
output
nisteaadddiiklooprrvz
input
abbbaaccca
ab
aca
output
ababacabcc
Note

In the third sample, this optimal solutions has three non-overlaping substrings equal to either b or c on positions 1 – 2 (ab), 3 – 4 (ab), 5 – 7 (aca). In this sample, there exist many other optimal solutions, one of them would be acaababbcc.

给定A,B,C串,其中A串(可能)包含B或C串,现在可以任意变换A串的位置,使A串中包含B串,C串的总个数最大。

所以贪心下啊,统计字母个数

#include <bits/stdc++.h>
using namespace std;
char a[],b[],c[];
int A[],B[],C[],S[];
int main() {
cin>>a>>b>>c;
int len=strlen(a);
for(int i=; i<len; i++)
A[a[i]-'a']++;
len=strlen(b);
for(int i=; i<len; i++)
B[b[i]-'a']++;
len=strlen(c);
for(int i=; i<len; i++)
C[c[i]-'a']++;
int ans,x,y;
ans=x=y=;
len=strlen(a)/strlen(b);
for(int i=; i<=len; i++) {
int ok=;
memcpy(S,A,sizeof(A));
for(int j=; j<; j++) {
if(B[j]*i>S[j]) {
ok=;
break;
}
S[j]-=B[j]*i;
}
if(!ok)
continue;
int t=;
for(int j=; j<; j++) {
if(C[j]==)
continue;
t=min(t,S[j]/C[j]);
}
if(t+i>ans) {
ans=t+i;
x=i;
y=t;
}
}
for(int i=; i<x; i++)
cout<<b;
for(int i=; i<y; i++)
cout<<c;
for(int i=; i<; i++) {
for(int j=; j<A[i]-x*B[i]-y*C[i]; j++)
cout<<char('a'+i);
}
cout<<endl;
return ;
}
 

Codeforces Round #307 (Div. 2)的更多相关文章

  1. 字符串处理/贪心 Codeforces Round #307 (Div. 2) B. ZgukistringZ

    题目传送门 /* 题意:任意排列第一个字符串,使得有最多的不覆盖a/b字符串出现 字符串处理/贪心:暴力找到最大能不覆盖的a字符串,然后在b字符串中动态得出最优解 恶心死我了,我最初想输出最多的a,再 ...

  2. 水题 Codeforces Round #307 (Div. 2) A. GukiZ and Contest

    题目传送门 /* 水题:开个结构体,rk记录排名,相同的值有相同的排名 */ #include <cstdio> #include <cstring> #include < ...

  3. Codeforces Round #307 (Div. 2) E. GukiZ and GukiZiana 分块

    E. GukiZ and GukiZiana Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/55 ...

  4. Codeforces Round #307 (Div. 2) C. GukiZ hates Boxes 贪心/二分

    C. GukiZ hates Boxes Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/551/ ...

  5. Codeforces Round #307 (Div. 2) B. ZgukistringZ 暴力

    B. ZgukistringZ Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/551/probl ...

  6. Codeforces Round #307 (Div. 2) A. GukiZ and Contest 水题

    A. GukiZ and Contest Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/551/ ...

  7. 「日常训练」ZgukistringZ(Codeforces Round #307 Div. 2 B)

    题意与分析(CodeForces 551B) 这他妈哪里是日常训练,这是日常弟中弟. 题意是这样的,给出一个字符串A,再给出两个字符串B,C,求A中任意量字符交换后(不限制次数)能够得到的使B,C作为 ...

  8. Codeforces Round #307 (Div. 2) D. GukiZ and Binary Operations (矩阵高速幂)

    题目地址:http://codeforces.com/contest/551/problem/D 分析下公式能够知道,相当于每一位上放0或者1使得最后成为0或者1.假设最后是0的话,那么全部相邻位一定 ...

  9. Codeforces Round #307 (Div. 2) E. GukiZ and GukiZiana (分块)

    题目地址:http://codeforces.com/contest/551/problem/E 将n平均分成sqrt(n)块,对每一块从小到大排序,并设置一个总体偏移量. 改动操作:l~r区间内,对 ...

  10. Codeforces Round #307 (Div. 2) D. GukiZ and Binary Operations 矩阵快速幂优化dp

    D. GukiZ and Binary Operations time limit per test 1 second memory limit per test 256 megabytes inpu ...

随机推荐

  1. 服务器 未能加载文件或程序集“XXXX”或它的某一个依赖项。试图加载格式不正确的程序。

    ,本人采用的第一种解决办法解决,已解决 问题2: 在同一个服务器上想要一个IP有两个网址,配置端口号,给新端口号开权限

  2. vue-cli脚手架(框架)

    一.创建vue项目 npm install vue-cli -g #-g全局 (sudo)npm install vue-cli -g #mac笔记本 vue-init webpack myvue # ...

  3. JS小游戏

    捕鱼达人 飞机大战游戏 详解javaScript的深拷贝 http://www.cnblogs.com/penghuwan/p/7359026.html

  4. cvLoadImage,cvCloneImage的内存泄露问题

    本文转自: http://hi.baidu.com/%C3%A8%D1%DB%D3%E3/blog/item/9d947e1b2b05555742a9adfd.html/cmtid/9872c2260 ...

  5. Objective-C Operators and Expressions

    What is an Expression? The most basic expression consists of an operator, two operands and an assign ...

  6. sublime快捷键mark

    Ctrl+D 选词 (反复按快捷键,即可继续向下同时选中下一个相同的文本进行同时编辑)Ctrl+G 跳转到相应的行Ctrl+J 合并行(已选择需要合并的多行时)Ctrl+L 选择整行(按住-继续选择下 ...

  7. 从零开发分布式数据库中间件 二、构建MyBatis的读写分离数据库中间件

    在上一节 从零开发分布式数据库中间件 一.读写分离的数据库中间件 中,我们讲了如何通过ThreadLocal来指定每次访问的数据源,并通过jdbc的连接方式来切换数据源,那么这一节我们使用我们常用的数 ...

  8. 查看进程lsof

    查看8000端口 lsof -i :8000 杀死进程 pkill -ns <pid>

  9. JS 、JQ 获取宽高总结 & JS中getBoundingClientRect的作用及兼容方案

    1.getBoundingClientRect的作用 getBoundingClientRect用于获取某个html元素相对于视窗的位置集合.   执行 object.getBoundingClien ...

  10. caffe修改需要的东西

    https://blog.csdn.net/zhaishengfu/article/details/51971768?locationNum=3&fps=1