AtCoder Beginner Contest 058 ABCD题
A - ι⊥l
Time limit : 2sec / Memory limit : 256MB
Score : 100 points
Problem Statement
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles beautiful if the tops of the poles lie on the same line, that is, b−a=c−b.
Determine whether the arrangement of the poles is beautiful.
Constraints
- 1≤a,b,c≤100
- a, b and c are integers.
Input
Input is given from Standard Input in the following format:
a b c
Output
Print YES if the arrangement of the poles is beautiful; print NO otherwise.
Sample Input 1
2 4 6
Sample Output 1
YES
Since 4−2=6−4, this arrangement of poles is beautiful.
Sample Input 2
2 5 6
Sample Output 2
NO
Since 5−2≠6−5, this arrangement of poles is not beautiful.
Sample Input 3
3 2 1
Sample Output 3
YES
Since 1−2=2−3, this arrangement of poles is beautiful.
题意:额。。
解法:额
#include<bits/stdc++.h>
using namespace std;
typedef long long ll; ll an[],am[];
int main()
{
int a,b,c;
cin>>a>>b>>c;
if(b-a==c-b)
{
cout<<"YES"<<endl;
}
else
{
cout<<"NO"<<endl;
}
return ;
}
B - ∵∴∵
Time limit : 2sec / Memory limit : 256MB
Score : 200 points
Problem Statement
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions.
You are given two strings O and E. O contains the characters at the odd-numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
Constraints
- O and E consists of lowercase English letters (
a-z). - 1≤|O|,|E|≤50
- |O|−|E| is either 0 or 1.
Input
Input is given from Standard Input in the following format:
O
E
Output
Print the original password.
Sample Input 1
xyz
abc
Sample Output 1
xaybzc
The original password is xaybzc. Extracting the characters at the odd-numbered positions results in xyz, and extracting the characters at the even-numbered positions results in abc.
Sample Input 2
atcoderbeginnercontest
atcoderregularcontest
Sample Output 2
aattccooddeerrbreeggiunlnaerrccoonntteesstt
题意:额
解法:额
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
string s1,s2,s3;
int main()
{
int num1=;
int num2=;
cin>>s1;
cin>>s2;
for(int i=;i<s1.size()+s2.size();i++)
{
if(i%)
{
cout<<s2[num1++];
}
else
{
cout<<s1[num2++];
}
}
cout<<endl;
return ;
}
C - 怪文書 / Dubious Document
Time limit : 2sec / Memory limit : 256MB
Score : 300 points
Problem Statement
Snuke loves "paper cutting": he cuts out characters from a newspaper headline and rearranges them to form another string.
He will receive a headline which contains one of the strings S1,…,Sn tomorrow. He is excited and already thinking of what string he will create. Since he does not know the string on the headline yet, he is interested in strings that can be created regardless of which string the headline contains.
Find the longest string that can be created regardless of which string among S1,…,Sn the headline contains. If there are multiple such strings, find the lexicographically smallest one among them.
Constraints
- 1≤n≤50
- 1≤|Si|≤50 for every i=1,…,n.
- Si consists of lowercase English letters (
a-z) for every i=1,…,n.
Input
Input is given from Standard Input in the following format:
n
S1
…
Sn
Output
Print the lexicographically smallest string among the longest strings that satisfy the condition. If the answer is an empty string, print an empty line.
Sample Input 1
3
cbaa
daacc
acacac
Sample Output 1
aac
The strings that can be created from each of cbaa, daacc and acacac, are aa, aac, aca, caa and so forth. Among them, aac, aca andcaa are the longest, and the lexicographically smallest of these three is aac.
Sample Input 2
3
a
aa
b
Sample Output 2
The answer is an empty string.
题意:选择字符串公共的字母,哪个字母出现次数最少就加进去,比如a在第一个字符串只出现了次,于是有aa
解法:模拟
#include <bits/stdc++.h> using namespace std;
string s[];
int n;
map<char,int>q;
string s1;
int main()
{
cin>>n;
for(int i=; i<n; i++)
{
cin>>s[i];
}
for(char i='a'; i<='z'; i++)
{
int flag=;
for(int j=; j<n; j++)
{
if(s[j].find(i)==-)
{
flag=;
}
}
if(flag==)
{
int minn=;
// cout<<i<<endl;
for(int j=; j<n; j++)
{
int cnt=;
for(int z=;z<s[j].size();z++)
{
if(s[j][z]==i)
{
cnt++;
}
}
minn=min(minn,cnt);
}
for(int j=;j<minn;j++)
{
s1+=i;
}
}
}
cout<<s1<<endl;
return ;
}
D - 井井井 / ###
Time limit : 2sec / Memory limit : 256MB
Score : 500 points
Problem Statement
On a two-dimensional plane, there are m lines drawn parallel to the x axis, and n lines drawn parallel to the y axis. Among the lines parallel to the x axis, the i-th from the bottom is represented by y=yi. Similarly, among the lines parallel to the y axis, the i-th from the left is represented by x=xi.
For every rectangle that is formed by these lines, find its area, and print the total area modulo 109+7.
That is, for every quadruple (i,j,k,l) satisfying 1≤i<j≤n and 1≤k<l≤m, find the area of the rectangle formed by the lines x=xi, x=xj, y=yk and y=yl, and print the sum of these areas modulo 109+7.
Constraints
- 2≤n,m≤105
- −109≤x1<…<xn≤109
- −109≤y1<…<ym≤109
- xi and yi are integers.
Input
Input is given from Standard Input in the following format:
n m
x1 x2 … xn
y1 y2 … ym
Output
Print the total area of the rectangles, modulo 109+7.
Sample Input 1
3 3
1 3 4
1 3 6
Sample Output 1
60
The following figure illustrates this input:

The total area of the nine rectangles A, B, ..., I shown in the following figure, is 60.

Sample Input 2
6 5
-790013317 -192321079 95834122 418379342 586260100 802780784
-253230108 193944314 363756450 712662868 735867677
Sample Output 2
835067060
题意:把这里面所有的正方形面积都加一次
解法: ∑∑(xi-xj)(yi-yj)1<=i<=n 1<=j<=m
把上面分开有:

于是...就出来了
#include<bits/stdc++.h>
//std::ios::sync_with_stdio(false);
using namespace std;
typedef long long ll;
ll n,m;
ll sx;
ll sy;
ll mod=1e9+;
ll x[],y[];
int main()
{
cin>>n>>m;
for(ll i=;i<=n;i++)
{
cin>>x[i];
}
for(ll i=;i<=m;i++)
{
cin>>y[i];
}
for(ll i=;i<=n;i++)
{
sx+=((i-)*x[i]-(n-i)*x[i])%mod;
sx%=mod;
}
for(ll i=;i<=m;i++)
{
sy+=((i-)*y[i]-(m-i)*y[i])%mod;
sy%=mod;
}
cout<<sx%mod*sy%mod<<endl;
return ;
}
AtCoder Beginner Contest 058 ABCD题的更多相关文章
- AtCoder Beginner Contest 068 ABCD题
A - ABCxxx Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement This contes ...
- AtCoder Beginner Contest 053 ABCD题
A - ABC/ARC Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement Smeke has ...
- AtCoder Beginner Contest 069 ABCD题
题目链接:http://abc069.contest.atcoder.jp/assignments A - K-City Time limit : 2sec / Memory limit : 256M ...
- AtCoder Beginner Contest 070 ABCD题
题目链接:http://abc070.contest.atcoder.jp/assignments A - Palindromic Number Time limit : 2sec / Memory ...
- AtCoder Beginner Contest 057 ABCD题
A - Remaining Time Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement Dol ...
- AtCoder Beginner Contest 051 ABCD题
A - Haiku Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement As a New Yea ...
- AtCoder Beginner Contest 052 ABCD题
A - Two Rectangles Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement The ...
- AtCoder Beginner Contest 054 ABCD题
A - One Card Poker Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement Ali ...
- AtCoder Beginner Contest 050 ABC题
A - Addition and Subtraction Easy Time limit : 2sec / Memory limit : 256MB Score : 100 points Proble ...
随机推荐
- mysql 环境变量之 group_concat_max_len
今天使用mysql group_concat()函数,对查询的数据进行字符串连接操作. 不过由于查询的结果较多,连接后的结果很长导致不能完全显示. 查询手册发现如下说明: (先说说group_conc ...
- MVC Controller构造器注入
UnityDependencyResolver 的标准写法 public class UnityDependencyResolver : IDependencyResolver { priva ...
- win7 64位安装vs2013 出现'System.AccessViolationException的错误
用管理员身份运行CMD,输入netsh winsock reset并回车(注意,必须是已管理员身份运行,这个重置LSP连接)
- android自己定义开关控件
近日在android项目要使用开关控件.可是android中自带的开关控件不太惬意,所以就打算通过自己定义View写一个开关控件 ios的开关控件当然就是我要仿照的目标. 先上图: waterma ...
- 获取IOS应用安装列表
原文转载至 http://blog.csdn.net/justinjing0612/article/details/8887747 转自鸟哥博客:http://blog.cnrainbird.com/ ...
- nodejs的request模块
request模块让http请求变的更加简单.(作为客户端,去请求.抓取另一个网站的信息) request的GitHub主页: https://github.com/request/request 最 ...
- HDU 6114 Chess 【组合数】(2017"百度之星"程序设计大赛 - 初赛(B))
Chess Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submi ...
- POJ3080 Blue Jeans —— 暴力枚举 + KMP / strstr()
题目链接:https://vjudge.net/problem/POJ-3080 Blue Jeans Time Limit: 1000MS Memory Limit: 65536K Total ...
- dede摘要长度,dedecms摘要限制,dedecms摘要字数
dede摘要长度,dedecms摘要限制,dedecms摘要字数 如果可以有效控制文章摘要的字数,那么就可以使得页面布局很灵活. 在Dedecms中,在列表页调用文章摘要的方法主要有: 1:[fiel ...
- hdu 2544 最短路 解题报告
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2544 题目意思:给出 n 个路口和 m 条路,每一条路需要 c 分钟走过.问从路口 1 到路口 n 需 ...