2 seconds
256 megabytes
standard input
standard output
Sometimes some words like "localization" or "internationalization"
are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters.
All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization»
will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
The first line contains an integer n (1 ≤ n ≤ 100).
Each of the following n lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Print n lines. The i-th line should contain the result
of replacing of the i-th word from the input data.
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <string>
#include <iostream>
using namespace std;
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
string s;
cin>>s;
int L=s.size();
if(L<=10) cout<<s<<endl;
else cout<<s[0]<<L-2<<s[L-1]<<endl;
}
return 0;
}
2 seconds
256 megabytes
standard input
standard output
A progress bar is an element of graphical interface that displays the progress of a process for this very moment before it is completed. Let's take a look at the following form of such a bar.
A bar is represented as n squares, located in line. To add clarity, let's number them with positive integers from 1 to n from
the left to the right. Each square has saturation (ai for
the i-th square), which is measured by an integer from 0 to k.
When the bar for some i (1 ≤ i ≤ n) is
displayed, squares 1, 2, ... , i - 1 has the saturation k,
squares i + 1, i + 2, ... , n has the saturation 0,
and the saturation of the square i can have any value from 0 to k.
So some first squares of the progress bar always have the saturation k. Some last squares always have the saturation 0.
And there is no more than one square that has the saturation different from 0 and k.
The degree of the process's completion is measured in percents. Let the process be t% completed. Then the following inequation is fulfilled:
An example of such a bar can be seen on the picture.
For the given n, k, t determine
the measures of saturation for all the squares ai of
the progress bar.
We are given 3 space-separated integers n, k, t (1 ≤ n, k ≤ 100, 0 ≤ t ≤ 100).
Print n numbers. The i-th of them should be equal
to ai.
#include <iostream>
using namespace std;
int main()
{
int n,k,t;
cin>>n>>k>>t;
//sum为Ai的和式的下界
int sum=t*n*k/100;
int cnt=sum/k;
for(int i=0;i<cnt;i++)//先填cnt个k
cout<<k<<" ";
if(cnt<n)
{
int p=sum-cnt*k;//补上差值
cout<<p;
for(int i=0;i<n-cnt-1;i++)//剩下的补0
cout<<" 0";
}
cout<<endl;
return 0;
}
3 seconds
256 megabytes
standard input
standard output
There are n knights sitting at the Round Table at an equal distance from each other. Each of them is either in a good or in a bad mood.
Merlin, the wizard predicted to King Arthur that the next month will turn out to be particularly fortunate if the regular polygon can be found. On all vertices of the polygon knights in a good
mood should be located. Otherwise, the next month will bring misfortunes.
A convex polygon is regular if all its sides have same length and all his angles are equal. In this problem we consider only regular polygons with at least 3 vertices, i. e. only nondegenerated.
On a picture below some examples of such polygons are present. Green points mean knights in a good mood. Red points mean ones in a bad mood.
King Arthur knows the knights' moods. Help him find out if the next month will be fortunate or not.
The first line contains number n, which is the number of knights at the round table (3 ≤ n ≤ 105).
The second line contains space-separated moods of all the n knights in the order of passing them around the table. "1" means that the knight is in a good
mood an "0" means that he is in a bad mood.
Print "YES" without the quotes if the following month will turn out to be lucky. Otherwise, print "NO".
#include <iostream>
#include <cstring>
using namespace std;
const int maxn =100000+5;
bool m[maxn];
int a[maxn];
int main()
{
int n,sum=0;
cin>>n;
for(int i=1; i<=n; i++)
{
cin>>a[i];
sum+=a[i];
}
if(n==sum)//全为1的特判
{
cout<<"YES"<<endl;
return 0;
}
bool flag=false;
for(int t=3; t<=n; t++)
if(n%t==0)//此处可以加个素数的判断
{
memset(m,0,sizeof(m));
for(int i=1; i<=n/t; i++)
{
if(a[i]==1)
{
for(int j=i; j<=n; j+=n/t)
if(a[i]+a[j]!=2) m[i]=1;
}
else m[i]=1;
}
int cnt=0;
for(int i=1; i<=n/t; i++)
if(!m[i]) cnt++;
if(cnt) flag=true;
}
if(flag) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
return 0;
}