标签:ack turn color str nbsp pre sync style href
题目连接:D. Rocket
题意:交互题,让你猜一个数字, 你每次询问一个数字,他会告诉你是大还是小,但是他回答不一定正确,有一个序列p,当p[i]为1时他会回答正确的,否则错误。询问次数<60,
题解:很明显的二分,但是p的长度有限<30我们可以通过一直猜1先把p[i]数组猜出来,然后在二分询问就可以了2^30>1e5
#include<bits/stdc++.h> #include<set> #include<cstdio> #include<iomanip> #include<iostream> #include<string> #include<cstring> #include<algorithm> #define pb push_back #define ll long long #define fi first #define se second #define PI 3.14159265 #define ls l,m,rt<<1 #define rs m+1,r,rt<<1|1 #define eps 1e-7 #define pii pair<int,int> typedef unsigned long long ull; const int mod=1e3+5; const ll inf=0x3f3f3f3f3f3f3f; const int maxn=1e5+5; using namespace std; int n,k,m; int op[100]; int main() { ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); cin>>m>>n; int x; for(int i=1;i<=n;i++) { cout<<1<<endl; cout.flush(); cin>>x; if(x==0)exit(0); else if(x==1)op[i]=1; else op[i]=0; } int l=1,r=m;int cn=1; while(l<=r) { int mid=(l+r)>>1; cout<<mid<<endl; cout.flush(); cin>>x; if(x==0)exit(0); if(op[cn]) { if(x==1)l=mid+1; else r=mid-1; } else { if(x!=1)l=mid+1; else r=mid-1; } cn++; if(cn>n)cn-=n; } return 0; }
Codeforces Round #499 (Div. 2)D. Rocket
标签:ack turn color str nbsp pre sync style href
原文地址:https://www.cnblogs.com/lhclqslove/p/9377124.html