Iterated Difference
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 786 Accepted Submission(s): 505
Problem Description
You
are given a list of N non-negative integers a(1), a(2), ... , a(N). You
replace the given list by a new list: the k-th entry of the new list is
the absolute value of a(k) - a(k+1), wrapping around at the end of the
list (the k-th entry of the new list is the absolute value of a(N) -
a(1)). How many iterations of this replacement are needed to arrive at a
list in which every entry is the same integer?
are given a list of N non-negative integers a(1), a(2), ... , a(N). You
replace the given list by a new list: the k-th entry of the new list is
the absolute value of a(k) - a(k+1), wrapping around at the end of the
list (the k-th entry of the new list is the absolute value of a(N) -
a(1)). How many iterations of this replacement are needed to arrive at a
list in which every entry is the same integer?
For example, let N = 4 and start with the list (0 2 5 11). The successive iterations are:
2 3 6 11
1 3 5 9
2 2 4 8
0 2 4 6
2 2 2 6
0 0 4 4
0 4 0 4
4 4 4 4
Thus, 8 iterations are needed in this example.
Input
The
input will contain data for a number of test cases. For each case,
there will be two lines of input. The first line will contain the
integer N (2 <= N <= 20), the number of entries in the list. The
second line will contain the list of integers, separated by one blank
space. End of input will be indicated by N = 0.
input will contain data for a number of test cases. For each case,
there will be two lines of input. The first line will contain the
integer N (2 <= N <= 20), the number of entries in the list. The
second line will contain the list of integers, separated by one blank
space. End of input will be indicated by N = 0.
Output
For
each case, there will be one line of output, specifying the case number
and the number of iterations, in the format shown in the sample output.
If the list does not attain the desired form after 1000 iterations,
print 'not attained'.
each case, there will be one line of output, specifying the case number
and the number of iterations, in the format shown in the sample output.
If the list does not attain the desired form after 1000 iterations,
print 'not attained'.
Sample Input
4
0 2 5 11
5
0 2 5 11 3
4
300 8600 9000 4000
16
12 20 3 7 8 10 44 50 12 200 300 7 8 10 44 50
3
1 1 1
4
0 4 0 4
0
0 2 5 11
5
0 2 5 11 3
4
300 8600 9000 4000
16
12 20 3 7 8 10 44 50 12 200 300 7 8 10 44 50
3
1 1 1
4
0 4 0 4
0
Sample Output
Case 1: 8 iterations
Case 2: not attained
Case 3: 3 iterations
Case 4: 50 iterations
Case 5: 0 iterations
Case 6: 1 iterations
Case 2: not attained
Case 3: 3 iterations
Case 4: 50 iterations
Case 5: 0 iterations
Case 6: 1 iterations
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
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 100
const int inf=0x7fffffff; //无限大
ll n[maxn];
ll n2[maxn];
int N;
int main()
{
int cas=;
int N;
while(cin>>N)
{
if(N==)
break;
for(int i=;i<N;i++)
{
cin>>n[i];
n2[i]=n[i];
}
ll ans=;
int flag=;
int kiss=;
for(int i=;i<N;i++)
{
if(n[i]!=n[(i+)%N])
{
kiss=;
break;
}
if(i==N-)
kiss=;
}
while(kiss==&&ans<=)
{ for(int i=;i<N;i++)
{
n[i]=fabs(n2[i]-n2[(i+)%N]);
}
for(int i=;i<N;i++)
{
n2[i]=n[i];
}
ans++;
for(int i=;i<N;i++)
{
if(n[i]!=n[(i+)%N])
{
kiss=;
break;
}
if(i==N-)
kiss=;
}
}
if(kiss==)
printf("Case %d: %d iterations\n",cas,ans);
else
printf("Case %d: not attained\n",cas);
cas++;
}
return ;
}