Kolakoski HDU多校签到题 6130

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 nth 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 integer n(1≤n≤107).

 

Output

For each test case:
A single line contains a nonnegative integer, denoting the answer.

 

Sample Input

212

 

Sample Output

12
题意即为  每连续的相同数字即为一组，ai表示第i组有ai个数。例如a3为2，表明第三组有两个数，即a4=a5=1（因为第二组的数为2，要与上一组的数不同）。因为a4为1，第四组就为一个数，即a6=2.
#include<iostream>using namespace std;// 1,2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1,2,1,1,2,1,2,2,1,1     Kolakoski数列int a[10000000]= {0};int main() {int t;a[1]=1;a[2]=2;a[3]=2;int m=3;          //记录当前ai为第m组for(int i=4; i<10000000; i++) {       if(a[m]==1) {              如果当前为单数组，保持与ai-1不同就行if(a[i-1]==1)a[i]=2;elsea[i]=1;m++;} else {                      如果当前为双数组，保持ai和ai+1与ai-1不同就行。if(a[i-1]==1)a[i+1]=a[i]=2;elsea[i+1]=a[i]=1;m++;i++;}}cin>>t;while(t--) {int n;cin>>n;cout<<a[n]<<endl;}return 0;}