交互题

Posted 2021-02-09 downrainsun

https://vjudge.net/contest/283149#problem/I 专题

Subway Pursuit

题意：猜火车的位置，

每次区间缩小到一定程度就可以用rand函数猜一个数字，猜对了就结束，猜错了之后，在这个区间火车可能会向左走K,向右走K。4k是一个人为订的范围，指区间变得这么大的时候就才一个数。大一点也行。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<ctime>
using namespace std;
typedef long long ll;

ll n, k;
char s[10];

int main() {
    srand(time(0));
    scanf("%lld%lld", &n, &k);
    ll l = 1, r = n;
    while(1){
        if(r - l <= (ll)4 * k){
            ll x = rand() % (r - l + 1) + l;
            printf("%lld %lld
", x, x);
            fflush(stdout);
            scanf("%s", s);
            if(s[0] == ‘Y‘ || s[0] == ‘B‘) return 0;
            else{
                l = max(1ll, l - k);
                r = min(n, r + k);
            }
        }
        else{
            ll mid = (l + r) / 2;
            printf("%lld %lld
", l, mid);
            fflush(stdout);
            scanf("%s", s);
            if(s[0] == ‘B‘) return 0;
            else if(s[0] == ‘Y‘){
                l = max(1ll, l - k);
                r = min(n, mid + k);
            }
            else{
                l = max(1ll, mid + 1 - k);
                r = min(n, r + k);
            }
        }
    }
    return 0;
}

 The hat

 尽量还是用cout

相邻的两个人相差1或-1（圆的）， 求那个人与对面那个人的树枝相等。必须是4K个人才有可能相等。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
const int maxm = 1e5 + 5;
typedef long long ll;
int  n;
int fun(int x) {
int l ,r;
cout << "? " << x <<endl;
cin >> l;
cout << "? " << x + n / 2 << endl;
cin >> r;
if(r - l == 0) {
cout << "! " << x <<endl;
    exit(0);
}
return r - l;
}


int main() {
cin >> n;
if(n % 4) {
cout << "! -1" << endl;
    return 0;
}
int l = 1, r = n / 2, mid;
while(l <= r) {
    mid = (l + r) / 2;
    ll vl = fun(l);
    ll ml = fun(mid);
    if(vl * ml < 0) {
        r = mid;
    }
    else {
        l = mid + 1;
    }
}
cout << "! -1" << endl;
return 0;
}