Codeforces1624 F. Interacdive Problem（交互，牛逼二分）

Posted 2022-02-14 live4m

题意：

解法：

在[1,n-1]中二分x的初始值mid,
check就是判断x的值是否>=mid,如何判断?
假设x=mid,之前累加的值为add,那么当前x就是nx=mid+add,
为了使得x越过n,我们的c至少为n-nx%n,就令c=n-nx%n
此时如果我们询问c,那么预期能得到(nx+c)/n.
如果实际拿到的确实是(nx+c)/n,那么说明x也确实>=mid.

二分过程中不断维护x可能存在的区间即可.

code：

#include<bits/stdc++.h>
// #define SYNC_OFF
#define int long long
#define ll long long
#define ull unsigned long long
//fast-coding
#define ST(x) x.begin()
#define ED(x) x.end()
#define RST(x) x.rbegin()
#define RED(x) x.end()
#define CL(x) x.clear();
#define all(a,n) a+1,a+1+n
#define ff(i,n) for(ll i=1;i<=n;i++)
#define rff(i,n) for(ll i=n;i>=1;i--)
#define fff(i,n) for(ll i=0;i<n;i++)
#define rfff(i,n) for(ll i=n-1;i>=0;i--)
#define SC(x) scanf("%s",x)
#define SL(x) strlen(x)
#define pss(a) push_back(a)
#define ps(a) push(a)
#define SZ(x) (int)x.size()
#define pee puts("");
#define eee putchar(' ');
#define re readdd()
#define pr(a) printtt(a)
int readdd()int x=0,f=1;char c=getchar();//
while(!isdigit(c)&&c!='-')c=getchar();
if(c=='-')f=-1,c=getchar();
while(isdigit(c))x=x*10+c-'0',c=getchar();
return f*x;
void printtt(int x)if(x<0)putchar('-'),x=-x;//
if(x>=10)printtt(x/10);putchar(x%10+'0');
int gcd(int a,int b)return b==0?a:gcd(b,a%b);//
int ppow(int a,int b,int mod)a%=mod;//
int ans=1%mod;while(b)if(b&1)ans=(long long)ans*a%mod;
a=(long long)a*a%mod;b>>=1;return ans;
bool addd(int a,int b)return a>b;
int lowbit(int x)return x&-x;
const int dx[4]=0,0,1,-1;
const int dy[4]=1,-1,0,0;
bool isdigit(char c)return c>='0'&&c<='9';
bool Isprime(int x)
    for(int i=2;i*i<=x;i++)if(x%i==0)return 0;
    return 1;

void ac(int x)if(x)puts("YES");else puts("NO");
//short_type 
#define VE vector<int>
#define PI pair<int,int>
//
using namespace std;
// const int mod=998244353;
const int mod=1e9+7;
const int maxm=2e6+5;
int add;//x已经累加的值
int n;
int ask(int x)
    add+=x;
    cout<<"+ "<<x<<endl;
    cin>>x;return x;

void out(int x)
    cout<<"! "<<x<<endl;

void solve()
    cin>>n;
    int l=1,r=n-1;
    int ans=-1;
    while(l<=r)
        int mid=(l+r)/2;//假设x初始值为mid
        int nx=mid+add;//那么当前x实际值为mid+add
        int c=n-nx%n;//为了使其越过一次n,我们至少要加上c
        int need=(nx+c)/n;//预期拿到(nx+c)/n
        int real=ask(c);//实际拿到了real
        if(real==need)//如果real>=need,符合预期,说明x在[mid,r]内
            ans=mid;
            l=mid+1;
        else//否则说明x在[l,mid-1]内
            r=mid-1;
        
    
    out(ans+add);//输出当前x

void Main()
    // #define MULTI_CASE
    #ifdef MULTI_CASE
    int T;cin>>T;while(T--)
    #endif
    solve();

void Init()
    #ifdef SYNC_OFF
    ios::sync_with_stdio(0);cin.tie(0);
    #endif
    #ifndef ONLINE_JUDGE
    freopen("../in.txt","r",stdin);
    freopen("../out.txt","w",stdout);
    #endif

signed main()
    Init();
    Main();
    return 0;

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