bzoj1044

题解：

第一问二分答案

第二问dp+单调队列

代码：

#include<bits/stdc++.h>
const int M=10007,N=50005;
typedef long long ll;
using namespace std;
int n,m,ans1,ans2,a[N],sum[N],f[2][N],q[N];
int jud(int x)
{
    int tmp=0,sum=0;
    for (int i=1;i<=n;i++)
     {
        sum+=a[i];
        if (sum>x){tmp++;sum=a[i];}
        if (tmp>m)return 0;
        if (a[i]>x)return 0;
     }
    return 1;
}
void solve1()
{
    int l=1,r=sum[n];
    while (l<=r)
     {
        int mid=(l+r)>>1;
        if (jud(mid)){ans1=mid;r=mid-1;}
        else l=mid+1;
     }
}
void solve2()
{
    f[0][0]=1;
    int pre,cur,tot;
    for (int i=1;i<=m;i++)
     {
        pre=i&1;cur=pre^1;
        int l=1,r=1;
        q[1]=0;tot=f[cur][0];
        for (int j=1;j<=n;j++)
         {
            while (l<=r&&sum[j]-sum[q[l]]>ans1)tot=(tot-f[cur][q[l++]]+M)%M;
            f[pre][j]=tot;q[++r]=j;
            tot=(tot+f[cur][j]+M)%M;
         }
        for (int j=n-1;j;j--)
         {
            if (sum[n]-sum[j]>ans1)break;
            ans2=(ans2+f[pre][j]+M)%M;
         }
        memset(f[cur],0,sizeof(f[cur]));
     }
    printf("%d\n",ans2);
}
int main()
{
    scanf("%d%d",&n,&m);
    for (int i=1;i<=n;i++)scanf("%d",&a[i]);
    for (int i=1;i<=n;i++)sum[i]=sum[i-1]+a[i];
    solve1();
    printf("%d ",ans1);
    solve2();
}