Codeforces Round #499 (Div. 2) D. Rocket题解

题目：

http://codeforces.com/contest/1011/problem/D

This is an interactive problem.

Natasha is going to fly to Mars. Finally, Natasha sat in the rocket. She flies, flies... but gets bored. She wishes to arrive to Mars already! So she decides to find something to occupy herself. She couldn't think of anything better to do than to calculate the distance to the red planet.

Let's define $\mathrm{xx as\; the\; distance\; to\; Mars.\; Unfortunately,\; Natasha\; does\; not\; know xx.\; But\; it\; is\; known\; that 1\le x\le m1\le x\le m,\; where\; Natasha\; knows\; the\; number mm.\; Besides, xx and mm are\; positive\; integers.}$

Natasha can ask the rocket questions. Every question is an integer $\mathrm{yy (1\le y\le m1\le y\le m).\; The\; correct\; answer\; to\; the\; question\; is -1-1,\; if x<yx<y, 00,\; if x=yx=y,\; and 11,\; if x>yx>y.\; But\; the\; rocket\; is\; broken —\; it\; does\; not\; always\; answer\; correctly.\; Precisely:\; let\; the\; correct\; answer\; to\; the\; current\; question\; be\; equal\; to tt,\; then,\; if\; the\; rocket\; answers\; this\; question\; correctly,\; then\; it\; will\; answer tt,\; otherwise\; it\; will\; answer -t-t.}$

In addition, the rocket has a sequence $\mathrm{pp of\; length nn.\; Each\; element\; of\; the\; sequence\; is\; either 00 or 11.\; The\; rocket\; processes\; this\; sequence\; in\; the\; cyclic\; order,\; that\; is 11-st\; element, 22-nd, 33-rd, \dots \dots , (n-1)(n-1)-th, nn-th, 11-st, 22-nd, 33-rd, \dots \dots , (n-1)(n-1)-th, nn-th, \dots \dots .\; If\; the\; current\; element\; is 11,\; the\; rocket\; answers\; correctly,\; if 00 —\; lies.\; Natasha\; doesn\text{'}t\; know\; the\; sequence pp,\; but\; she\; knows\; its\; length — nn.}$

You can ask the rocket no more than $\mathrm{6060 questions.}$

Help Natasha find the distance to Mars. Assume, that the distance to Mars does not change while Natasha is asking questions.

Your solution will not be accepted, if it does not receive an answer $\mathrm{00 from\; the\; rocket\; (even\; if\; the\; distance\; to\; Mars\; is\; uniquely\; determined\; by\; the\; already\; received\; rocket\text{'}s\; answers).}$

Input

The first line contains two integers $\mathrm{mm and nn (1\le m\le 1091\le m\le 109, 1\le n\le 301\le n\le 30) —\; the\; maximum\; distance\; to\; Mars\; and\; the\; number\; of\; elements\; in\; the\; sequence pp.}$

Interaction

You can ask the rocket no more than $\mathrm{6060 questions.}$

To ask a question, print a number $\mathrm{yy (1\le y\le m1\le y\le m)\; and\; an\; end-of-line\; character,\; then\; do\; the\; operation flush and\; read\; the\; answer\; to\; the\; question.}$

If the program reads $\mathrm{00,\; then\; the\; distance\; is\; correct\; and\; you\; must\; immediately\; terminate\; the\; program\; (for\; example,\; by\; calling exit(0)).\; If\; you\; ignore\; this,\; you\; can\; get\; any\; verdict,\; since\; your\; program\; will\; continue\; to\; read\; from\; the\; closed\; input\; stream.}$

If at some point your program reads $-2-2 as\; an\; answer,\; it\; must\; immediately\; end\; (for\; example,\; by\; calling exit(0)).\; You\; will\; receive\; the\; "Wrong\; answer"\; verdict,\; and\; this\; will\; mean\; that\; the\; request\; is\; incorrect\; or\; the\; number\; of\; requests\; exceeds 6060.\; If\; you\; ignore\; this,\; you\; can\; get\; any\; verdict,\; since\; your\; program\; will\; continue\; to\; read\; from\; the\; closed\; input\; stream.$

If your program's request is not a valid integer between $-231-231 and 231-1231-1 (inclusive)\; without\; leading\; zeros,\; then\; you\; can\; get\; any\; verdict.$

You can get "Idleness limit exceeded" if you don't print anything or if you forget to flush the output.

To flush the output buffer you can use (after printing a query and end-of-line):

• fflush(stdout) in C++;
• System.out.flush() in Java;
• stdout.flush() in Python;
• flush(output) in Pascal;
• See the documentation for other languages.

Hacking

Use the following format for hacking:

In the first line, print $\mathrm{33 integers m,n,xm,n,x (1\le x\le m\le 1091\le x\le m\le 109, 1\le n\le 301\le n\le 30) —\; the\; maximum\; distance\; to\; Mars,\; the\; number\; of\; elements\; in\; the\; sequence pp and\; the\; current\; distance\; to\; Mars.}$

In the second line, enter $\mathrm{nn numbers,\; each\; of\; which\; is\; equal\; to 00 or 11 —\; sequence pp.}$

The hacked solution will not have access to the number $\mathrm{xx and\; sequence pp.}$

Example

input

5 2
1
-1
-1
1
0

output

1
2
4
5
3
题意：互动问题，告诉你一个数m，让你在1到m之间猜一个数x，你每次询问系统会告诉你三种答案：0（猜对了），1（猜小了），-1（猜大了），但是系统告诉你的不一定是真话（/TДT)/,
系统有一个长度为n(n<=30)的由0和1组成的数组,你知道n但不知道数组具体内容，系统回答时会按照该数组顺序来，如果a[i]==1,他会正确回答，如果a[i]==0，他会回答相反数.
即第一次询问系统根据a[1]来确定是否给出正确的回答，第2次看a[2]，……第n次看a[n],第n+1次看a[1]，循环.
在不超过60次询问下问出正确答案（即系统回答0则正确）;
思路：
前n次询问边界（1或m）确定该数组，之后就用二分的问法去寻找答案.
代码:

#include<bits/stdc++.h>
#define fi first
#define se second
#define INF 0x3f3f3f3f
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define pqueue priority_queue
#define NEW(a,b) memset(a,b,sizeof(a))
const double pi=4.0*atan(1.0);
const double e=exp(1.0);
const int maxn=3e6+8;
typedef long long LL;
typedef unsigned long long ULL;
//typedef pair<LL,LL> P;
const LL mod=1e9+7;
using namespace std;
int a[110];
int main(){
    int m,n;
    cin>>m>>n;
    int x;
    int cnt=0;
    while(1){
        cout<<1<<endl;
        cin>>x;
        if(x==0){
            return 0;
        }
        if(x==-1){
            a[cnt++]=-1;
        }
        else if(x==1){
            a[cnt++]=1;
        }
        if(cnt==n){
            break;
        }
    }
    int l=1;
    int r=m+1;
    int i=0;
    while(1){
        int mid=(l+r)/2;
        cout<<mid<<endl;
        cin>>x;
        if(x==0){
            return 0;
        }
        if(a[i]==-1){
            x=-x;
        }
        i++;
        if(i==n){
            i=0;
        }
        if(x==-1){
            r=mid;
        }
        else if(x==1){
            l=mid+1;
        }
    }
}