「SHOI2015」自动刷题机

/*
有理有据的二分答案
因为在过程中最多减到零  所以n越小显然就能刷更多的题 

无解时就是无论如何也无法得到k ， 这个特判一下即可 
*/ 

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#define M 100010
#define ll long long 
#define inf 100000000000ll
using namespace  std;
ll read()
{
    ll nm = 0, f = 1;
    char c = getchar();
    for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
    return nm * f;
}
ll note[M], n, k;

ll check(ll x)
{
    ll sum = 0, ans = 0;
    for(int i = 1; i <= n; i++)
    {
        sum += note[i];
        if(sum >= x) ans++, sum = 0;
        if(sum < 0) sum = 0;
    }
    return ans;
}

int main()
{
    n = read(), k = read();
    for(int i = 1; i <= n; i++) note[i] = read();
    ll l = 1, r = inf, ln, rn;
    while(l + 1 < r)
    {
        ll mid = (l + r) >> 1;
        if(check(mid) <= k) r = mid;
        else l = mid;
    }
    if(check(l) == k) r = l;
    ln = r;
    l = 1, r = inf;
    while(l + 1 < r)
    {
        ll mid = (l + r) >> 1;
        if(check(mid) >= k) l = mid;
        else r = mid;
    }
    if(check(r) == k) l = r;
    rn = l;
    if(ln > rn || check(ln) != k) return puts("-1");
    cout << ln << " " << rn << "
";
    return 0;
}