忘开long long了居然没WA
二分答案，枚举判断看最后需要的月份数是否小于等于要求的即可
#include<iostream>
#include<cstdio>
using namespace std;
const int N=100005;
int n,m,a[N],l,r,ans;
int read()
{
int r=0,f=1;
char p=getchar();
while(p>'9'||p<'0')
{
if(p=='-')
f=-1;
p=getchar();
}
while(p>='0'&&p<='9')
{
r=r*10+p-48;
p=getchar();
}
return r*f;
}
bool ok(int mid)
{
int con=1,sum=0;
for(int i=1;i<=n;i++)
{
if(sum+a[i]>mid)
sum=a[i],con++;
else
sum+=a[i];
}
return con<=m;
}
int main()
{
n=read(),m=read();
for(int i=1;i<=n;i++)
a[i]=read(),l=max(l,a[i]),r+=a[i];
while(l<=r)
{
int mid=(l+r)>>1;
if(ok(mid))
r=mid-1,ans=mid;
else
l=mid+1;
}
printf("%d
",ans);
return 0;
}