HDU 4791 Alice's Print Service 水二分

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Alice's Print Service

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1084    Accepted Submission(s): 240

Problem Description

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 ≤ 10⁵ ). The second line contains 2n integers s₁, p₁ , s₂, p₂ , ..., s_n, p_n (0=s₁ < s₂ <
... < s_n ≤ 10⁹ , 10⁹ ≥ p₁ ≥ p₂ ≥ ... ≥ 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 q₁ .. q_m (0 ≤ q_i ≤ 10⁹ ) 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

 

Source

field=problem&key=2013+Asia+Changsha+Regional+Contest+&source=1&searchmode=source" style="color:rgb(26,92,200); text-decoration:none">2013 Asia Changsha Regional Contest

 

先逆序求出最小。然后二分查找要求的位置。

//484MS	2556K
#include<stdio.h>
#include<algorithm>
#define inf 1e13
using namespace std;
__int64 s[100007];
__int64 x[100007],y[100007];
int main()
{
    //freopen("in.txt","r",stdin);
    int t;
    scanf("%d",&t);
    while(t--)
    {
        __int64 mi;
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            scanf("%I64d %I64d",&x[i],&y[i]);
        for(int i=1;i<=n;i++)s[i]=x[i]*y[i];
        s[n+1]=inf;
        mi=s[n+1];
        for(int i=n;i>=1;i--)
        {
            mi=min(s[i],mi);
            s[i]=mi;
        }
        while(m--)
        {
            __int64 q,minn;
            scanf("%I64d",&q);
            int pos=lower_bound(x+1,x+n+1,q)-x;//在first和last中的前闭后开区间进行二分查找。返回大于或等于val的第一个元素位置。假设全部元素都小于val。则返回last的位置
            minn=min(q*y[pos],s[pos+1]);
            printf("%I64d\n",minn);
        }
    }
    return 0;
}