GYM 100685 J【交互题】

俄罗斯的人经常出一些交互题，比如强制离线之类的题目
这题是二分+交互
对于每一盏灯i，我们假设前面的灯位置都排好了位置，那么就二分那些这一盏灯所在的位置，询问的次数是nlog(n)次。
另外如果死循环的话，那就是没有方案，设置一个cnt上限来判断死循环。

//      whn6325689//      Mr.Phoebe//      http://blog.csdn.net/u013007900#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath>#include <functional>#include <numeric>#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define eps 1e-9#define PI acos(-1.0)#define INF 0x3f3f3f3f#define LLINF 1LL<<62#define speed std::ios::sync_with_stdio(false);typedef long long ll;typedef unsigned long long ull;typedef long double ld;typedef pair<ll, ll> pll;typedef complex<ld> point;typedef pair<int, int> pii;typedef pair<pii, int> piii;typedef vector<int> vi;#define CLR(x,y) memset(x,y,sizeof(x))#define CPY(x,y) memcpy(x,y,sizeof(x))#define clr(a,x,size) memset(a,x,sizeof(a[0])*(size))#define cpy(a,x,size) memcpy(a,x,sizeof(a[0])*(size))#define debug(a) cout << #a" = " << (a) << endl;#define debugarry(a, n) for (int i = 0; i < (n); i++) { cout << #a"[" << i << "] = " << (a)[i] << endl; }#define mp(x,y) make_pair(x,y)#define pb(x) push_back(x)#define lowbit(x) (x&(-x))#define MID(x,y) (x+((y-x)>>1))#define ls (idx<<1)#define rs (idx<<1|1)#define lson ls,l,mid#define rson rs,mid+1,rtemplate<class T>inline bool read(T &n){    T x = 0, tmp = 1;    char c = getchar();    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();    if(c == EOF) return false;    if(c == '-') c = getchar(), tmp = -1;    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();    n = x*tmp;    return true;}template <class T>inline void write(T n){    if(n < 0)    {        putchar('-');        n = -n;    }    int len = 0,data[20];    while(n)    {        data[len++] = n%10;        n /= 10;    }    if(!len) data[len++] = 0;    while(len--) putchar(data[len]+48);}//-----------------------------------int n;vector<int> ans;bool judge(int x,int y){    string str;    cout<<1<<" "<<x<<" "<<y<<endl;    cout.flush();    cin>>str;    if(str[0]=='Y') return 1;    return 0;}int main(){    bool flag = true;    int l, r, mid, pos, cnt;    scanf("%d",&n);    ans.push_back(1);    for (int i = 2; i <= n; i++)    {        cnt = 0;        l = 0; r = i-2; pos = 0;        while (l <= r)        {            mid = (l+r)>>1; cnt++;            if (cnt > 100000)            {                flag = false;                break;            }            if (judge(ans[mid], i))            {                l = mid+1;                pos = mid+1;            }            else r = mid-1;        }        if (!flag) break;        ans.insert(ans.begin()+pos, i);    }    printf("0 ");    if (flag)    {        for (int i = 0; i <= n-2; i++) printf("%d ", ans[i]);        printf("%d\n", ans[n-1]);    }    else    {        for (int i = 0; i <= n-2; i++) printf("0 ");        printf("0\n");    }    //system("pause");    return 0;}