1、题意：题意很简洁吧，就不概括了

2、分析：我思考了半天，我猜答案满足单调。。。没敢写，看了题解去问Claris为啥单调，Claris一句话“ 因为n越大明显不可能做更多题 ”，后来没找到反例我也放弃了

满足单调的话就二分答案咯

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
#define M 200010
#define LL long long

inline int read(){
    char ch = getchar(); int x = 0, f = 1;
    while(ch < '0' || ch > '9'){
        if(ch == '-') f = -1;
        ch = getchar();
    }
    while('0' <= ch && ch <= '9'){
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
} 

int a[M];
int n, m;

inline int check(LL x){
    int cnt = 0;
    LL now = 0;
    for(int i = 1; i <= n; i ++){
        now += a[i];
        now = max(0ll, now);
        if(now >= x) cnt ++, now = 0;
    }
    return cnt;
}

int main(){
    n = read(), m = read();
    for(int i = 1; i <= n; i ++) a[i] = read();
    LL l = 1, r = 10000000000000, t1 = 1, t2 = 10000000000000;
    while(l <= r){
        LL mid = (l + r) / 2;
        if(check(mid) >= m) l = (t1 = mid) + 1;
        else r = mid - 1;
    }
    l = 1, r = 10000000000000;
    while(l <= r){
        LL mid = (l + r) / 2;
        if(check(mid) <= m) r = (t2 = mid) - 1;
        else l = mid + 1;
    }
    if(check(t2) != m || check(t1) != m){
        printf("-1");
        return 0;
    }
    printf("%lld %lld", t2, t1);
    return 0;
}