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/*0 3, 4 6 8 9 11 13 15,16 18 20 21 23 25 27,28
30 32 33 35 37 39,40 42 44 45 47 49 51,
52 54 56 57 59 61 63,64 66 68 69 71 73 75
,76 78 80 81 83 85 87,88*/#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<string.h>
#include<string>
#include<cmath>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<bits/stdc++.h>
typedef long long ll;
using namespace std;/* 本来想着要判断各个字母的前缀，后来发现还可以跨组别来求相隔的半音，所以应该想到要直接用map把这个按键的名字当做一个下标存下来 现代音乐用七个英文字母 C D E F G A B（或其小写）来标记音名。
这七个不同高低的音，其相邻音之间的音高距离，有半音和全音之分，其中E与F 和 B与C 之间为半音关系，其余相邻音之间为全音关系。
现代钢琴键盘上的全音和半音
相邻两个琴键（包括白健和黑键）构成半音，相隔一个琴键的两键之间构成全音。*/
int main(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int case1;map<string,int>shu;//map按key的ascii进行排序，而不是按输入顺序 cin>>case1;string name[52]={"A_2","B_2","C_1","D_1","E_1","F_1","G_1","A_1","B_1","C","D","E","F","G","A","B","c","d","e","f","g","a","b","c^1","d^1","e^1","f^1","g^1","a^1","b^1","c^2","d^2","e^2","f^2","g^2","a^2","b^2","c^3","d^3","e^3","f^3","g^3","a^3","b^3","c^4","d^4","e^4","f^4","g^4","a^4","b^4","c^5"};int zhi[52]={0,3,4,6,8,9,11,13,15,16,18,20,21,23,25,27,28,30,32,33,35,37,39,40,42,44,45,47,49,51,52,54,56,57,59,61,63,64,66,68,69,71,73,75,76,78,80,81,83,85,87,88};for(int i=0;i<52;i++){shu.insert(make_pair(name[i],zhi[i]));}
//  for(map<string,int>::iterator it=shu.begin();it!=shu.end();it++){
//      cout<<it->first<<" "<<it->second<<endl;
//  }while(case1--) {string s1,s2;cin>>s1>>s2;cout<<abs(shu[s2]-shu[s1])<<endl;}
}

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#include<iostream>
#include<algorithm>
typedef long long ll;
using namespace std;
const int maxn=110;
struct ani{ll x;ll y;}an[maxn];bool cmp1(ani a,ani b){return a.x>b.x;
}bool cmp2(ani a,ani b){return a.y<b.y;}int main(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int T,n,m;cin>>T;while(T--){ll ans=0,ans1=0;cin>>n>>m;for(int i=0;i<n;i++){cin>>an[i].x>>an[i].y;}sort(an,an+n,cmp1);for(int j=0;j<m;j++)ans+=an[j].x;cout<<ans<<" ";sort(an,an+n,cmp2);for(int j=0;j<m;j++)ans1+=an[j].y;cout<<ans1<<endl;}}

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#include<iostream>
typedef long long ll;
using namespace std;
int main(){int T;ll a,b,c;ll ans;cin>>T;while(T--){cin>>a>>b>>c;if(((a-b)<1)&&c<=3500||(a-b)==0){cout<<"impossible"<<endl;}else{ans=3500/(a-b);if((3500%(a-b))>0)ans++;if(ans>c){cout<<"impossible"<<endl;}else{cout<<ans<<endl;}}} }/*
这一题一定要预判a-b的值是否为0
比如说：
①除以零
②数组越界：int a[3]; a[10000000]=10;
③指针越界：int * p; p=(int *)malloc(5 * sizeof(int)); *(p+1000000)=10;
④使用已经释放的空间：int * p; p=(int *)malloc(5 * sizeof(int));free(p); *p=10;
⑤数组开得太大，超出了栈的范围，造成栈溢出：int a[100000000];
*/

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#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<map>
typedef long long ll;
using namespace std;
int main(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int case1;cin>>case1;while(case1--){string s,tmp;cin>>s;if(s=="Rough"){cin>>tmp;cout<<"It will be a hard work!"<<endl; }else if(s=="Sunny"){cout<<"Stay"<<endl;  }else{cout<<"Let us go!"<<endl;}}
}

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#include<iostream>
typedef long long ll;
using namespace std;
const int maxn=1e7+5;
int arr[maxn]={2,3,3};//数据开到1e4以上就不能在main函数里面定义数组了，否则程序会
//直接蹦 ，这一题打表来做
int main(){int T,case1;int cur=1;int last=2;int i=3; while(i<maxn){if(arr[last]==3){if(arr[cur]==3){arr[i]=2;arr[i+1]=2;arr[i+2]=2;i=i+3;  last+=3;}else{arr[i]=2;arr[i+1]=2;i=i+2;last+=2;}}else{if(arr[cur]==3){arr[i]=3;arr[i+1]=3;arr[i+2]=3;i=i+3;  last+=3;}else{arr[i]=3;arr[i+1]=3;i=i+2;last+=2;}}cur++;}//  for(int i=0;i<10000;i++)cout<<arr[i]<<" ";cin>>T;while(T--){cin>>case1;cout<<arr[case1-1]<<endl;    }}
解法二：
#include<bits/stdc++.h>
using namespace std;
const int xa=1e7+5;
int a[xa];
int main()
{int p;cin>>p;a[0]=2;int pan=1;int shu=0;int i=0;while(shu<xa){if(a[i]==2){int t=2;while(t--){if(pan%2==1){a[++shu]=3;}else{a[++shu]=2;}}pan++;}if(a[i]==3){int t=3;while(t--){if(pan%2==1){a[++shu]=3;}else{a[++shu]=2;}}pan++;}i++;}int x;while(p--){cin>>x;cout<<a[x-1]<<endl;}}