A - Alice's Print Service
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit 
Status 
Practice 
HDU 4791

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 ≤ 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  思路:稍微拿多一点可能更便宜

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <iostream>
#define INF 1e18
using namespace std;

const int maxn=100500;

typedef long long LL;

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

int main()
{
	int T,n,m;
	LL mid,mmin,res;
	scanf("%d",&T);
	while(T--){
		scanf("%d%d",&n,&m);
		for(int i=0;i<n;i++){
			scanf("%I64d%I64d",&s[i],&p[i]);
			h[i]=s[i]*p[i];
		}
		mmin=INF;
		for(int i=n-1;i>=0;i--){
			mmin=min(s[i]*p[i],mmin);
			h[i]=mmin;
		}
		for(int kk=0;kk<m;kk++){
			scanf("%I64d",&mid);
			if(mid>=s[n-1]) printf("%I64d\n",mid*p[n-1]);
			else{
                int t=upper_bound(s,s+n,mid)-s;
                res=mid*p[t-1];
                res=min(res,h[t]);
                printf("%I64d\n",res);
			}
		}
	}
	return 0;
}

转载于:https://www.cnblogs.com/xuesu/p/3979851.html

原文链接：https://blog.csdn.net/weixin_30342827/article/details/95660364

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