嘟嘟嘟
这题就比较水了，毕竟只评了个蓝。
想一下发现满足单调性，所以可以二分找最大值。
但是最小值怎么办？刚开始我很zz的以为只要把判断条件从大于等于改成小于等于就行了，后来发现根本不对。
想了想因为最小值和最大值之间一定是一段答案为\(k\)的区间，所以可以二分找最小值：如果当前答案不等于\(m\)，就向右二分，否则向左二分。
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e5 + 5;
inline ll read()
{
ll ans = 0;
char ch = getchar(), last = ' ';
while(!isdigit(ch)) last = ch, ch = getchar();
while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
if(last == '-') ans = -ans;
return ans;
}
inline void write(ll x)
{
if(x < 0) x = -x, putchar('-');
if(x >= 10) write(x / 10);
putchar(x % 10 + '0');
}
int n, m;
ll a[maxn], Max;
In bool judge1(ll x)
{
ll cnt = 0; ll tot = 0;
for(int i = 1; i <= n; ++i)
{
tot = max(1LL * 0, tot + a[i]);
if(tot >= x) ++cnt, tot = 0;
}
return cnt >= m;
}
In bool judge2(ll x)
{
ll cnt = 0; ll tot = 0;
for(int i = 1; i <= n; ++i)
{
tot = max(1LL * 0, tot + a[i]);
if(tot >= x) ++cnt, tot = 0;
}
return cnt == m;
}
int main()
{
// freopen("1.in", "r", stdin);
n = read(), m = read();
for(int i = 1; i <= n; ++i) a[i] = read();
ll L = 0, R = 1e14;
while(L < R)
{
ll mid = (L + R + 1) >> 1;
if(judge1(mid)) L = mid;
else R = mid - 1;
}
if(!L || !judge2(L)) {puts("-1"); return 0;}
Max = L;
L = 1, R = Max;
while(L < R)
{
ll mid = (L + R) >> 1;
if(judge2(mid)) R = mid;
else L = mid + 1;
}
write(L), space, write(Max), enter;
return 0;
}