D - Finding Zero

D - Finding Zero

#includeusing namespace std;#define rep(i,l,r) for(int i=(l);i<=(r);i++)#define per(i,l,r) for(int i=(l);i>=(r);i--)#define ll long long#define pii pair#define mset(s,t) memset(s,t,sizeof(t))#define mcpy(s,t) memcpy(s,t,sizeof(t))#define fir first#define pb push_back#define sec secondinline int read () { int x = 0, f = 0; char ch = getchar(); while (!isdigit(ch)) f |= (ch=='-'), ch= getchar(); while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar(); return f?-x:x;}template void print(T x) { if (x < 0) putchar('-'), x = -x; if (x >= 10) print(x/10); putchar(x % 10 + '0');}const int N = 1e5 + 10;ll n, x, y;ll get (int x, int y, int z) { cout << "? " << x << " " << y << ' ' << z << endl; fflush(stdout); int k; cin >> k; return k;}void solve() { cin >> n; ll x, y, z; x = 1, y = 2, z = 3; int last = get (x, y, z); for (int i = 4; i<= n; i++) { int t = get (x, y, i); if (t >= last) { last = t; z = i; } } for (int i = 1; i<= n; i ++) { if (i ==x || i == z || i == y) continue; int t = get (x, i, z); if (t >= last) { last = t; y = i; } } int id = - 1; for (int i = 1; i <= n; i++) if (i == x || i == y || i == z) continue; else { id = i; break; } int k = get (x, y, id); if (k == last) { cout << "!" << " " << x << " " << y << endl; } else { k = get (x, id, z); if (k == last) { cout << "!" << " " << x << " " << z << endl; } else { cout << "!" << " " << y << " " << z << endl; } }}int main () { int t; cin >> t; while (t --) solve(); return 0;}

这次比赛卡在了B题，没想对方向，之前是是整个数来看，但是太麻烦了，应该设法简单点，正解是按照奇偶性思考，因为题目保证，最终只有一个能到达y，通过考虑最后一位异或操作和加操作是等价的。因为都是改变了奇偶性 找到一个性质，找到最大值最小值，然后再通过一些询问就可以得到答案

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