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最新推荐文章于 2019-08-10 13:10:08 发布

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本文介绍了一个编程挑战，任务是计算给定整数列表通过迭代差分操作达到所有元素相同所需的步数。具体步骤为：每次迭代将每个元素替换为其与下一个元素的绝对差值，最后一个元素与第一个元素形成闭环。

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Iterated Difference

Time Limit: 1000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

Submit Status

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?

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.

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'.

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

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

Source

The 2011 Rocky Mountain Regional Contest

#include<bits/stdc++.h>
using namespace std;
int a[30];
int main()
{
int n,ca=1;
while(scanf("%d",&n)&&n){

int k=1;
for(int i=0;i<n;i++) {scanf("%d",&a[i]);if(i!=0&&a[i]!=a[i-1]) k=0;}
if(n==1) k=1;
//if(k!=1 && k%2==1) k=2;
a[n]=a[0];
//for(int j=0;j<=n;j++) printf("%d ",a[j]);
// printf("\n");

int cnt=0;
while(k==0){
k=1;

// for(int j=0;j<=n;j++) printf("%d ",a[j]);
// printf("\n");

int i;
for(i=0;i<n;i++){
a[i]=abs(a[i+1]-a[i]);
if(i!=0&&a[i]!=a[i-1]) k=0;
}
a[n]=a[0];

cnt++;
if(cnt>1000) { k=2; break; }
}
if(k==1)
printf("Case %d: %d iterations\n",ca++,cnt);
else if(k==2)
printf("Case %d: not attained\n",ca++);
}

return 0;
}