Codeforces1301B. Motarack's Birthday

题意是说给你一串数组，其中-1代表未知，求相邻两个数之差的绝对值最小，-1可以由k赋值，先考虑-1的情况，把k解出来，转换一下，就是绝对值之差最小情况，|k-a|，|k-b|，|k-c|，要使最大的最小，也就是在两个端点找，|k-min|,|k-max|，由数形结合可知，当k为min与max的中点时最小，由此算出k再扫一遍数组即可

#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) ((x)&(-x))
typedef long long LL;
 
const int maxm = 1e5+5;
const int INF = 0x3f3f3f3f;
 
int buf[maxm], n;
 
void run_case() {
    cin >> n;
    int Max = -1, Min = INF;
    for(int i = 0; i < n; ++i) cin >> buf[i];
    for(int i = 0; i < n; ++i) {
        if(buf[i] == -1) {
            if(i > 0 && buf[i-1] != -1) Min = min(Min, buf[i-1]), Max = max(Max, buf[i-1]);
            if(i < n-1 && buf[i+1] != -1) Min = min(Min, buf[i+1]), Max = max(Max, buf[i+1]);
        }
    }
    int mid = (Max+Min)>>1, ans = 0;
    for(int i = 0; i < n-1; ++i) {
        int a = buf[i], b = buf[i+1];
        if(a == -1) a = mid;
        if(b == -1) b = mid;
        ans = max(ans, abs(a-b));
    }
    cout << ans << " " << mid << "
";
}
 
int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    //cout.setf(ios_base::showpoint);cout.precision(10);
    int t; cin >> t;
    while(t--)
    run_case();
    cout.flush();
    return 0;
}