知识点:二分
这个是最小化最大值的贪心,是可行解是从大到小也就是从右到左判断的,
#include <bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define mk make_pair
#define sz(x) ((int) (x).size())
#define all(x) (x).begin(), (x).end()
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pa;
const int N = 1e5 + 5;
int n, m, a[N];
bool check(int x) {
int cnt = 1;
int rec = a[0];
if (a[0] > x) return false;
for (int i = 1; i < n; i++) {
if (a[i] > x) return false;
if (rec + a[i] <= x) rec += a[i];
else { cnt++; rec = a[i]; }
if (cnt > m) return false;
}
return true;
}
void solve(int l, int r) {
while (l < r) {
int mid = (l + r) >> 1;
if (check(mid)) r = mid;
else l = mid + 1;
}
cout << l << endl;
}
int main() {
cin >> n >> m;
for (int i = 0; i < n; i++) cin >> a[i];
solve(0, 1000000000);
return 0;
}