Steph is extremely obsessed with “sequence problems” that are usually seen on magazines: Given the sequence 11, 23, 30, 35, what is the next number? Steph always finds them too easy for such a genius like himself until one day Klay comes up with a problem and ask him about it.

Given two integer sequences {ai} and {bi} with the same length n, you are to find the next n numbers of {ai}: an+1…a2n. Just like always, there are some restrictions on an+1…a2n: for each number ai, you must choose a number bk from {bi}, and it must satisfy ai≤max{aj-j│bk≤j<i}, and any bk can’t be chosen more than once. Apparently, there are a great many possibilities, so you are required to find max{∑2nn+1ai} modulo 109+7 .

Now Steph finds it too hard to solve the problem, please help him.

The input contains no more than 20 test cases.
For each test case, the first line consists of one integer n. The next line consists of n integers representing {ai}. And the third line consists of n integers representing {bi}.
1≤n≤250000, n≤a_i≤1500000, 1≤b_i≤n.

 

For each test case, print the answer on one line: max{∑2nn+1ai} modulo 109+7。

 

 

 #include <cstdio>

 #include <cstring>

 #include <vector>

 #include <algorithm>

 using namespace std;

 const int N=*+;

 const int M= 1e9+;

 typedef long long ll;

 struct node{

     int l,r;

     int Max;

     ll  sum;

 }tree[*N];

 int n;

 int a[N],b[N];

 void pushup(int i){

     tree[i].sum=(tree[*i].sum+tree[*i+].sum)%M;//别忘了mod运算

     tree[i].Max=max(tree[*i].Max,tree[*i+].Max);

 }

 void build(int l,int r,int i){

     if(i>*n)   return ;

     tree[i].l=l;

     tree[i].r=r;

     if(tree[i].l==tree[i].r) {

         if(l>n){

             tree[i].sum=;

             tree[i].Max=;

         }else{

             tree[i].sum=a[l];

             tree[i].Max=a[l]-l;//MAX存的是a[j]-j;

         }

         return ;

     }

     int mid=(l+r)/;

     build(l,mid,*i);

     build(mid+,r,*i+);

     pushup(i);//回溯更新父节点

 }

 void update(ll v,int x,int i){

     if(tree[i].l==tree[i].r){

         tree[i].sum=v;

         tree[i].Max=v-x;

         return ;

     }

     int mid=(tree[i].l+tree[i].r)/;

     if(x<=mid)  update(v,x,*i);

     else update(v,x,*i+);

     pushup(i);

 }

 int query(int l,int r,int i){

     if(tree[i].l==l && tree[i].r==r)    return tree[i].Max;

     int mid=(tree[i].l+tree[i].r)/;

     if(r<=mid)  return query(l,r,*i);

     else if(l>mid)  return query(l,r,*i+);

     else if(l<=mid && r>mid)    return max(query(l,mid,*i),query(mid+,r,*i+));

     return -;

 }

 int main(){

     while(scanf("%d",&n)!=EOF){

         int pre=;

         for(int i=;i<=n;i++)   scanf("%d",&a[i]);

         for(int i=;i<n;i++)    scanf("%d",&b[i]);

         build(,*n,);

         sort(b,b+n);

         for(int i=n+;i<=*n;i++){

             int x,y;

             int bg,ed=i-;

             x=bg=b[pre++];//排序完直接按顺序取b数组，保证了不会重复使用

             y=query(bg,ed,);

             a[i]=max(x,y);

             update(a[i],i,);

         }

         printf("%lld\n",tree[].sum);//tree[3]保存的是n+1…2*n的节点信息

     }

     return ;

 }