标签：ros memset code ext \n traints stdin case +=
Problem Statement
This is an interactive task.
Snuke has a favorite positive integer, N. You can ask him the following type of question at most 64 times: "Is n your favorite integer?" Identify N.
Snuke is twisted, and when asked "Is n your favorite integer?", he answers "Yes" if one of the two conditions below is satisfied, and answers "No" otherwise:
Here, str(x) is the decimal representation of x (without leading zeros) as a string. For example, str(123)= 123 and str(2000) = 2000. Strings are compared lexicographically. For example, 11111 < 123 and 123456789 < 9.
Constraints
Input and Output
Write your question to Standard Output in the following format:
? n
Here, n must be an integer between 1 and 1018 (inclusive).
Then, the response to the question shall be given from Standard Input in the following format:
ans
Here, ans is either Y or N. Y represents "Yes"; N represents "No".
Finally, write your answer in the following format:
! n
Here, n=N must hold.
Judging
TLE.WA).Sample
Below is a sample communication for the case N=123:
| Input | Output |
|---|---|
? 1 |
|
Y |
|
? 32 |
|
N |
|
? 1010 |
|
N |
|
? 999 |
|
Y |
|
! 123 |
题意：有一个1~1e9的数字，让你去猜。
你可以做最多64个询问，每一个询问评测机会根据这个规定来返回信息。
Snuke is twisted, and when asked "Is n your favorite integer?", he answers "Yes" if one of the two conditions below is satisfied, and answers "No" otherwise:
思路：
先确定这个数多少位，然后每一位用二分去得到具体这一位的数字。
我们从1到10，再100 ，1000， 每一次*10的去询问，就可以得到这个数字的位数。
如果我们问10，返回Y，问100，返回N，那么这一位是2位数，可以对照规定自己琢磨为什么。
知道多少位只有，我们每一个位置
int mid;
int l=0;
int r=9;
这样二分。
这里我利用了多一位的数一定n>N来一直进入第二个条件来询问的，
比如 是二位数，我问的时候问三位数，已知位放再数字中，未知位放0，询问位通过二分进行变化，
那么一定进入第二个条件，根据字典序的关系，来确定这一位的数字是几。
对于1和100这种1和1后面只有0的数，我们通过询问全是9的数来特判。
细节见代码：
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <strstream> #include <stack> #include <map> #include <set> #include <vector> #include <iomanip> #define ALL(x) (x).begin(), (x).end() #define rt return #define dll(x) scanf("%I64d",&x) #define xll(x) printf("%I64d\n",x) #define sz(a) int(a.size()) #define all(a) a.begin(), a.end() #define rep(i,x,n) for(int i=x;i<n;i++) #define repd(i,x,n) for(int i=x;i<=n;i++) #define pii pair<int,int> #define pll pair<long long ,long long> #define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define MS0(X) memset((X), 0, sizeof((X))) #define MSC0(X) memset((X), ‘\0‘, sizeof((X))) #define pb push_back #define mp make_pair #define fi first #define se second #define eps 1e-6 #define gg(x) getInt(&x) #define db(x) cout<<"== [ "<<x<<" ] =="<<endl; using namespace std; typedef long long ll; ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} ll lcm(ll a,ll b){return a/gcd(a,b)*b;} ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;} inline void getInt(int* p); const int maxn=1000010; const int inf=0x3f3f3f3f; /*** TEMPLATE CODE * * STARTS HERE ***/ int query(string ans) { char x; cout<<"? "<<ans<<endl; cin>>x; if(x==‘Y‘) { return 1; }else { return 0; } } void solve() { string temp="9"; string ans="1"; while(!query(temp)) { temp+="9"; ans+="0"; } cout<<"! "<<ans<<endl; } int main() { //freopen("D:\\common_text\\code_stream\\in.txt","r",stdin); //freopen("D:\\common_text\\code_stream\\out.txt","w",stdout); string ans="1"; string res; while(1) { cout<<"? "<<ans<<endl; cin>>res; if(res[0]==‘Y‘) { ans+="0"; }else { break; } if(ans.length()>11) { solve(); return 0; } } int len=ans.length(); string temp; rep(i,0,len-1) { int mid; int l=0; int r=9; while(l<=r) { mid=(l+r)>>1; ans[i]=‘0‘+mid; if(query(ans)) { r=mid-1; }else { l=mid+1; temp=ans; } } ans=temp; } temp.pop_back(); int num=temp.length(); stringstream ss; ss.clear(); ss<<temp; ll fans; ss>>fans; fans++; cout<<"! "<<fans<<endl; return 0; } inline void getInt(int* p) { char ch; do { ch = getchar(); } while (ch == ‘ ‘ || ch == ‘\n‘); if (ch == ‘-‘) { *p = -(getchar() - ‘0‘); while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) { *p = *p * 10 - ch + ‘0‘; } } else { *p = ch - ‘0‘; while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) { *p = *p * 10 + ch - ‘0‘; } } }
Awkward Response AtCoder - 2656 ( 二分+交互题)
标签：ros memset code ext \n traints stdin case +=
原文地址：https://www.cnblogs.com/qieqiemin/p/10815578.html
