hdu 4791 二分

Alice's Print Service

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1241 Accepted Submission(s): 287

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

12 30 20 100 100 99 100

Sample Output

010001000

Source

2013 Asia Changsha Regional Contest 

求当前数”所在区间“和“当前数后面区间两数的乘积”的最小值，用二分解决

#include<stdio.h>#define N 100005typedef __int64 ll;ll a[N],b[N],sum[N];ll find(ll l,ll r,ll k){    ll ans,mid;    while(l<=r)    {        mid=(l+r)>>1;        if(a[mid]<=k)        {            ans=mid;            l=mid+1;        }        else            r=mid-1;    }    return ans;}int main(){    __int64 t,n,m,i,q,k,ans,c;    scanf("%I64d",&t);    while(t--)    {        scanf("%I64d%I64d",&n,&m);        for(i=1;i<=n;i++)            scanf("%I64d%I64d",&a[i],&b[i]);        sum[n]=a[n]*b[n];        for(i=n-1;i>=1;i--)        {            c=a[i]*b[i];            sum[i]=sum[i+1]<c ? sum[i+1] : c;        }        while(m--)        {            scanf("%I64d",&q);            k=find(1,n,q);            //printf("k=%I64d\n",k);            ans=b[k]*q;            //printf("ans=%I64d  sum=%I64d\n",ans,sum[k+1]);            if(ans>sum[k+1]&&k+1<=n)                ans=sum[k+1];            printf("%I64d\n",ans);        }    }    return 0;}