EOJ Monthly 2019.2 A. 回收卫星

题目传送门

题意：

 你可以询问一个三维坐标，机器会告诉你这个坐标在不在目标圆中，

 并且（0，0，0）是一定在圆上的，叫你求出圆心坐标

思路：

 因为（0，0，0）一定在圆上，所以我们可以把圆心分成3个坐标轴上

 就是求x的时候，其他坐标都为0，而且可以确定一个在x的正半轴，一个

在负半轴，所以我们可以二分求出

另外注意数据（l+r）是数据会超过long long

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll INF=4e9;

void ask(ll x,ll y,ll z)
{
    printf("0 %lld %lld %lld\n",x,y,z);
    fflush(stdout);
}
ll xx0()
{
    ll l=-INF,r=0;
    ll op,ans;
    while(l<=r)
    {
        ll mid=(l+r)>>1;
        ask(mid,0,0);
        scanf("%d",&op);
        if(op==0)
            l=mid+1;
        else r=mid-1,ans=mid;
    }
    return ans;
}
ll xx1()
{
    ll l=0,r=INF;
    ll op,ans;
    while(l<=r)
    {
        ll mid=l+r>>1;
        ask(mid,0,0);
        scanf("%d",&op);
        if(op==0)
            r=mid-1;
        else l=mid+1,ans=mid;
    }
    return ans;
}
ll yy0()
{
    ll l=-INF,r=0;
    ll op,ans;
    while(l<=r)
    {
        ll mid=l+r>>1;
        ask(0,mid,0);
        scanf("%d",&op);
        if(op==0)
            l=mid+1;
        else r=mid-1,ans=mid;
    }
    return ans;
}
ll yy1()
{
    ll l=0,r=INF;
    ll op,ans;
    while(l<=r)
    {
        ll mid=l+r>>1;
        ask(0,mid,0);
        scanf("%d",&op);
        if(op==0)
            r=mid-1;
        else l=mid+1,ans=mid;
    }
    return ans;
}
ll zz0()
{
    ll l=-INF,r=0;
    ll op,ans;
    while(l<=r)
    {
        ll mid=l+r>>1;
        ask(0,0,mid);
        scanf("%d",&op);
        if(op==0)
            l=mid+1;
        else r=mid-1,ans=mid;
    }
    return ans;
}
ll zz1()
{
    ll l=0,r=INF;
    ll op,ans;
    while(l<=r)
    {
        ll mid=l+r>>1;
        ask(0,0,mid);
        scanf("%d",&op);
        if(op==0)
            r=mid-1;
        else l=mid+1,ans=mid;
    }
    return ans;
}
int main()
{
   ll x0,y0,x1,y1,z0,z1;
   x0=xx0();
   x1=xx1();
   y0=yy0();
   y1=yy1();
   z0=zz0();
   z1=zz1();
   printf("1 %lld %lld %lld\n",(x0+x1)/2,(y0+y1)/2,(z0+z1)/2);

    return 0;
}

View Code

#include<bits/stdc++.h>

#define ll long long

using namespace std;

const ll INF=2e9;

ll a[5];

void ask(int tp,int m) {

    for(int i=1; i<=3; i++)a[i]=0;

    a[tp]=m;

    printf("0 %lld %lld %lld\n",a[1],a[2],a[3]);

    fflush(stdout);

}

ll go(int tp) {

    ll l=0,r=INF,mid,ans1=0,ans2=0;

    int op;

    while(l<=r) {

        mid=l+(r-l)/2;

        ask(tp,mid);

        scanf("%d",&op);

        if(op)l=mid + 1,ans1 = mid;

        else r=mid - 1;

    }

    l=-INF,r=0;

    while(l<=r) {

        mid=l+(r-l)/2;

        ask(tp,mid);

        scanf("%d",&op);

        if(op)r=mid-1,ans2 = mid;

        else l=mid+1;

    }

    return (ans1+ans2)/2;

}

int main() 

{

    ll x,y,z;

    x=go(1);

    y=go(2);

    z=go(3);

    printf("1 %lld %lld %lld\n",x,y,z);

    return 0 ;

}

View Code