【LOJ】#2569. 「APIO2016」最大差分

题解

第一个子任务直接询问最大最小，每次可以问出来两个，再最大最小-1再问两个，最多问(frac{N + 1}{2})次就还原出了序列

第二个子任务由于差分肯定会大于等于(lceil frac{mx - mn}{N - 1} ceil)
那么我们直接把序列最大最小按照这个值分块，只用两个块之间的差分大小来更新答案

代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('
')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 1000005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 + c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
void MinMax(int64 s,int64 t,int64 &mn,int64 &mx) {
    putchar('?');space;out(s);space;out(t);enter;
    fflush(stdout);
    read(mn);read(mx);
}
int64 inf = 1e18;
int64 findGap(int T,int N) {
    int64 mn,mx;
    int64 ans = 0;
    if(T == 1) {
	MinMax(0,inf,mn,mx);
	int64 ls = mn,lt = mx;
	for(int i = 1 ; i <= (N + 1) / 2 - 1 ; ++i) {
	    if(ls + 1 > lt - 1) break;
	    MinMax(ls + 1,lt - 1,mn,mx);
	    if(mn == -1) break;
	    ans = max(mn - ls,ans);
	    ans = max(lt - mx,ans);
	    ls = mn,lt = mx;
	}
	ans = max(ans,lt - ls);
    }
    else {
	MinMax(0,inf,mn,mx);
	int64 len = (mx - mn - 1) / (N - 1) + 1;
	int64 a = mn,m2 = mx;
	int64 pre = mn;
	ans = len;
	for(int i = 1 ; i <= N - 1 ; ++i) {
	    int64 b;
	    b = min(a + len,m2);
	    MinMax(a,b,mn,mx);
	    if(mn != -1) {
		ans = max(ans,mn - pre);
		pre = mx;
	    }
	    if(a > m2 - len - 1) break;
	    a = a + len + 1;
	}
    }
    return ans;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    int T,N;
    read(T);read(N);
    int64 x = findGap(T,N);
    putchar('!');space;out(x);enter;fflush(stdout);
    return 0;
}