HDU 4791 Alice's Print Service (2013长沙现场赛，二分)

Alice's Print Service

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1855    Accepted Submission(s): 454

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

2013 Asia Changsha Regional Contest

【题意】：

给出一个打印的分段函数，求打印每一个query的张数，最少花多少钱

【解题思路】：

若pi <= x < pi+1 ；最小花费即Min｛pi*x, 用i后面的方案花的最少的钱｝;

逆序记录一下Min[i]，为用 i 以后的方案（多打印一些）所花费的最少的钱；；

WA了一次，longlong下忘记把inf赋为0x3f3f3f3f3f3f3f3f~

最近做水题都不太细心，，反省

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<vector>
 #include<map>
 #include<queue>
 #include<algorithm>
 #define LL long long
 #define maxn 110000
 #define inf 0x3f3f3f3f3f3f3f3f
 #define IN freopen("in.txt","r",stdin);
 using namespace std;

 LL p[maxn],s[maxn];
 LL tol[maxn];

 int main(int argc, char const *argv[])
 {
     //IN;

     int t;scanf("%d",&t);
     while(t--)
     {
         int n,m;
         scanf("%d %d",&n,&m);

         for(int i=;i<n;i++){
             scanf("%lld %lld",&s[i],&p[i]);
             tol[i] = s[i] * p[i];
         }

         LL Min[maxn];
         fill(Min,Min+maxn,inf);
         for(int i=n-;i>=;i--){
             Min[i]=min(tol[i],Min[i+]);
         }

         while(m--)
         {
             LL x;scanf("%lld",&x);
             LL ans = inf;
             int i = lower_bound(s,s+n,x) - s;
             if(s[i]==x) ans = min(p[i]*s[i],Min[i]);
             else ans = min(Min[i],p[i-]*x);

             printf("%lld\n",ans);
         }
     }

     return ;
 }