正题

题目链接:https://www.luogu.org/problem/P4343

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题目大意

$n$个操作每个操作加几行代码或减几行代码，若代码积累到$x$行就自动删除所有代码并切掉一道题。

已知道切掉了$k$题，求最大和最小的$x$

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解题思路

因为$x$和切题的数量单调，所以可以二分答案即可。

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$code$

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long 
using namespace std;
const ll N=100010;
ll n,k,a[N],sum,maxs,l,r;
bool check(ll x)
{ll now=0,z=0; for(ll i=1;i<=n;i++){now+=a[i];now=max(now,0ll);if(now>=x) z++,now=0;}return (z>=k);
}
bool check2(ll x)
{ll now=0,z=0; for(ll i=1;i<=n;i++){now+=a[i];now=max(now,0ll);if(now>=x) z++,now=0;}return (z<=k);
}
bool checks(ll x)
{ll now=0,z=0; for(ll i=1;i<=n;i++){now+=a[i];now=max(now,0ll);if(now>=x) z++,now=0;}return (z==k);
}
int main()
{scanf("%lld%lld",&n,&k);for(ll i=1;i<=n;i++)scanf("%lld",&a[i]),sum+=a[i],maxs=max(maxs,sum);l=1;r=(1e9)*n;while(l<=r){ll mid=(l+r)/2;if(check2(mid)) r=mid-1;else l=mid+1;}if(checks(l))printf("%lld ",l);else {printf("-1");return 0;}l=1;r=(1e9)*n;while(l<=r){ll mid=(l+r)/2;if(check(mid)) l=mid+1;else r=mid-1;}if(checks(r))printf("%lld",r);else printf("-1");
}