題目連結：傳送門

題目大意：

　　有n棵植物依次放在1-n，機器人從0出發澆水，每棵植物被澆水時di += ai，求澆m次水後min｛di|1 ≤ i ≤ n｝的最大值。（澆水時必須往左或往右走一步，落腳點為澆水點）

思路：

　　若已知答案mid（滑稽臉）：可以貪心地從左到右澆水，反覆給di和di+1澆水，直到當前的di ≥ mid。。。O(n)。。。既然可以O(n)驗證答案，那就開心地二分掉了。

程式碼：

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const
 
 int MAX_N = 1e5 + 5;

int N;
ll M;
ll a[MAX_N];
ll d[MAX_N];

bool judge(ll mid) {
    for (int i = 1; i <= N; i++) d[i] = 0;
    ll cnt = 0, jumped = 0;
    for (int i = 1; i <= N; i++) {
        if (d[i] >= mid) {
            jumped++;
            continue;
        }
        cnt += jumped; jumped = 0
 
;
        ll mul = (mid-d[i])/a[i] + ((mid-d[i])%a[i] > 0);
        cnt += mul*2-1;
        d[i] += mul * a[i];
        if (i+1 <= N)
            d[i+1] += (mul-1) * a[i+1];
    }
    return cnt <= M;
}

int main()
{
    int T;
    cin >> T;
    while (T--) {
        scanf("%d%lld", &N, &M);
        ll _max 
 
= 0;
        for (int i = 1; i <= N; i++) {
            scanf("%lld", a+i);
            _max = max(_max, a[i]);
        }

        ll l = 0, r = _max*M;
        ll ans = 0;
        while (l <= r) {
            ll mid = (l+r) >> 1;
            if (judge(mid)) {
                ans = max(ans, mid);
                l = mid+1;
            }
            else
                r = mid-1;
        }
        cout << ans << endl;
    }
    return 0;
}
/*
2
4 8
3 2 6 6
3 9
10 10 1
*/