Alice is providing print service, while the pricing doesn't seem to be reasonable, so people using her print service found some tricks to save money.
For example, the price when printing less than 100 pages is 20 cents per page, but when printing not less than 100 pages, you just need to pay only 10 cents per page. It's easy to figure out that if you want to print 99 pages, the best choice is to print an extra blank page so that the money you need to pay is 100 × 10 cents instead of 99 × 20 cents.
Now given the description of pricing strategy and some queries, your task is to figure out the best ways to complete those queries in order to save money.

Input

The first line contains an integer T (≈ 10) which is the number of test cases. Then T cases follow.
Each case contains 3 lines. The first line contains two integers n, m (0 < n, m ≤ 10 5 ). The second line contains 2n integers s 1, p 1 , s 2, p 2 , ..., s n, p n (0=s 1 < s 2 < ... < s n ≤ 10 9 , 10 9 ≥ p 1 ≥ p 2 ≥ ... ≥ p n ≥ 0).. The price when printing no less than s i but less than s i+1 pages is p i cents per page (for i=1..n-1). The price when printing no less than s n pages is p n cents per page. The third line containing m integers q 1 .. q m (0 ≤ q i ≤ 10 9 ) are the queries.

Output

For each query q i, you should output the minimum amount of money (in cents) to pay if you want to print q i pages, one output in one line.

Sample Input

1
2 3
0 20 100 10
0 99 100

Sample Output

0
1000
1000

给出不同范围内的价格，求最少需要花多少钱

对于每一个询问，二分找到其所在区间，然后比较打印正好的页数的价格和多于要求页数的价格（即区间端点），求一个最小值

暴力妥妥的超时，然后套上了线段树，WA了好多次

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f3f3f3fLL;
const int N=1e5+20;
ll a[N],b[N],T[4*N+50];
void pushup(int k)
{
    T[k]=min(T[k*2],T[k*2+1]);
}
void build(int l,int r,int k)
{
    if(l==r)
    {
        T[k]=a[l]*b[l];
        return ;
    }
    int mid=(l+r)/2;
    build(l,mid,k*2);
    build(mid+1,r,k*2+1);
    pushup(k);
}
ll Q(int L,int R,int x,int y,int k)
{
    if(L==x&&R==y)///要找到准确的对应的区间
        return T[k];
    int mid=(L+R)/2;
    if(y<=mid)
    {
        return Q(L,mid,x,y,k*2);///需要查询的区间都在同一侧，可以直接返回
    }
    else if(x>mid)
    {
        return Q(mid+1,R,x,y,k*2+1);
    }
    else
        return min(Q(L,mid,x,mid,k*2),Q(mid+1,R,mid+1,y,k*2+1));///将需要查询的区间由mid分割成两部分
}
int main()
{
    ll n,m;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(T,inf,sizeof(T));
        scanf("%lld%lld",&n,&m);
        for(int i=1; i<=n; i++)
            scanf("%lld%lld",&a[i],&b[i]);
        build(1,n,1);
        ll p;
        while(m--)
        {
            scanf("%lld",&p);
            ll ans;
            if(p>=a[n])
                ans=p*b[n];
            else if(p==0)
                ans=0;
            else
            {
                ll id=lower_bound(a+1,a+n+1,p)-a;
//                cout<<id<<'\n';
                if(id>=1&&id<=n)
                {
                    if(a[id]>p)
                        id--;
//                cout<<ans<<'\n';
                    ans=min(b[id]*p,Q(1,n,id+1,n,1));
                }
                else
                    ans=p*b[n];
            }
            printf("%lld\n",ans);
        }
    }
    return 0;
}

RMQ 

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e5+10;
int n,m;
ll a[N],b[N];
ll f[N][20];///f[i][j]表示从i位起2^j个数中的最大值，即【i,i+(1<<j)-1】
void ST_prework()
{
    for(int i=1; i<=n; i++)
        f[i][0]=a[i]*b[i];
    ll imax=(ll)(log(n*1.0)/log(2.0));
    for(int i=1; i<=imax; i++)
    {
        for(int j=1; j+(1<<i)-1<=n; j++) ///右端点为j+(1<<i)-1,-1是因为要包含j自己
            f[j][i]=min(f[j][i-1],f[j+(1<<(i-1))][i-1]);
    }
}
ll ST_query(ll l,ll r)///求(l,r)中的最大值
{
    ll k=(ll)(log((l-r+1)*1.0)/log(2.0));///区间长度r-l+1
    return min(f[l][k],f[r-(1<<k)+1][k]);///第1个区间：[l,l+(1<<k)-1];第2个区间：(r,(1<<k)+1~r)
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1; i<=n; i++)
            scanf("%lld%lld",&a[i],&b[i]);
        ST_prework();
//        cout<<a[1]<<' '<<a[2]<<' '<<b[1]<<' '<<b[2]<<'\n';
        while(m--)
        {
            ll q,ans;
            scanf("%lld",&q);
            ll id=lower_bound(a+1,a+n+1,q)-a;
//            cout<<id<<'\n';
            if(id>n||a[id]>q)
                id--;
            ans=q*b[id];
//            cout<<ans<<'\n';
            if(id<n)
                ans=min(ans,ST_query(id+1,n));
            cout<<ans<<'\n';
        }
    }
    return 0;
}