zoj 3726

Alice's Print Service

---

Time Limit: 2 Seconds      Memory Limit: 65536 KB

---

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. ThenT cases follow.

Each case contains 3 lines. The first line contains two integers n, m (0 <n, m ≤ 10⁵). The second line contains 2n integerss₁, 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 isp_i cents per page (for i=1..n-1). The price when printing no less thans_n pages is p_n cents per page. The third line containingm 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 printq_i pages, one output in one line.

Sample Input

12 30 20 100 100 99 100

Sample Output

010001000

题意：给出n种方案，当要购买的量达到si以上时，可用每件pi的价格购买。例如要购买99件物品，给出的方案是购买0件以上每件20和购买100件以上每件10，显然是多购买一件的方案才可以使开销最小。

思路：假设购买num件，一开始我以为只要二分找到一个最接近num但比num小或等于的s.i，若s.i对应的价格是p.i，答案必然是min（num*p.i，s.i+1*p.i+1）。但后来发现有这么一种情况样例，无论够买什么答案都是0，这让我想到后面的可能比前面的小，所以一开始就对数据进行预处理，从后往前推，存放s.i后面的所有s.i*p.i的最小值，答案必定就是这个值跟num*p.i之一

1

5 5

0 20 100 15 200 10 300 5 500 0

0 100 200 300 500

#include<stdio.h>#include<stdlib.h>#include<string.h>struct sw{long long s,p;}E[100005];long long dp[100005];long long min(long long a,long long b){return (a<b?a:b);}int main(void){int n,m,T;long long num,ans,ans1,ans2;scanf("%d",&T);while(T--){scanf("%d%d",&n,&m);for(int i=0;i<n;i++)scanf("%lld%lld",&E[i].s,&E[i].p);dp[n-1]=E[n-1].s*E[n-1].p;for(int i=n-2;i>=0;i--)dp[i]=min(E[i].s*E[i].p,dp[i+1]);while(m--){scanf("%lld",&num);int l,r,mid;l=0,r=n-1;while(l<r){mid=(l+r+1)/2;if(num>=E[mid].s)l=mid;else r=mid-1;} ans1=num*E[l].p;if(l==n-1) ans2=ans1;else ans2=dp[l+1];ans=min(ans1,ans2);printf("%lld\n",ans);}}return 0;}