hdu 6130 Kolakoski(多校联赛)
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Kolakoski
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 0 Accepted Submission(s): 0
Problem Description
This is Kolakosiki sequence: 1,2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1…… . This sequence consists of 1 and 2 , and its first term equals 1 . Besides, if you see adjacent and equal terms as one group, you will get 1,22,11,2,1,22,1,22,11,2,11,22,1…… . Count number of terms in every group, you will get the sequence itself. Now, the sequence can be uniquely determined. Please tell HazelFan its n th element.
Input
The first line contains a positive integer T(1≤T≤5) , denoting the number of test cases.
For each test case:
A single line contains a positive integern(1≤n≤107) .
For each test case:
A single line contains a positive integer
Output
For each test case:
A single line contains a nonnegative integer, denoting the answer.
A single line contains a nonnegative integer, denoting the answer.
Sample Input
212
Sample Output
12
水题一道 但是学到了一个新的高深而难懂数列
ac代码:
#include<bits/stdc++.h>#define inf 10000000using namespace std;int a[10000000+5];int ke(){ a[1]=1,a[2]=2,a[3]=2; int sum=3; for(int i=3;i<=inf;i++) { if(a[i]==2) { if(a[sum]==2) { a[sum+1]=a[sum+2]=1; } if(a[sum]==1) { a[sum+1]=a[sum+2]=2; } } if(a[i]==1) { if(a[sum]==2) { a[sum+1]=1; } if(a[sum]==1) { a[sum+1]=2; } } sum=sum+a[i]; if(sum>=inf) break; }}int main(){ ke(); int t,n; while(scanf("%d",&t)!=EOF) { while(t--) { scanf("%d",&n); printf("%d\n",a[n]); } } return 0;}
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