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  • The 2013 ACM-ICPC Asia Changsha Regional Contest

    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. 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, ..., sn, pn (0=s1 < s2 < ... < sn ≤ 109, 109 ≥ p1 ≥ p2 ≥ ... ≥ pn ≥ 0). The price when printing no less than si but less than si+1 pages is pi cents per page (for i=1..n-1). The price when printing no less than sn pages is pn cents per page. The third line containing m integers q1 .. qm (0 ≤ qi ≤ 109) are the queries.

    Output

    For each query qi, you should output the minimum amount of money (in cents) to pay if you want to print qi pages, one output in one line.

    Sample Input

    1
    2 3
    0 20 100 10
    0 99 100

    Sample Output

    0
    1000
    1000

    #include <iostream>
    #include <string>
    #include <string.h>
    #include <map>
    #include <stdio.h>
    #include <algorithm>
    #include <vector>
    #define Min(a,b) ((a)<(b)?(a):(b))
    using namespace std ;
    const int Max_N = 100008 ;
    typedef long long LL ;
    LL dp[Max_N] ;
    LL S[Max_N] ,P[Max_N] ;
    int N ,M ;
    
    int find_id(LL x){
       int Left = 1 ;
       int Right = N ;
       int mid ;
       int ans = -1;
       while(Left<=Right){
           mid = (Left + Right)>>1 ;
           if(S[mid] >= x){
               Right = mid - 1 ;
               ans = mid ;
           }
           else
               Left = mid + 1 ;
       }
       return ans ;
    }
    
    int main(){
        int T ;
        LL x;
        scanf("%d",&T) ;
        while(T--){
           scanf("%d%d",&N,&M) ;
           for(int i = 1 ;i <= N ;i++)
              scanf("%lld%lld",&S[i],&P[i]) ;
           dp[N] = S[N] * P[N] ;
           for(int i = N-1 ;i >= 1 ;i--){
                dp[i] = Min(S[i]*P[i],dp[i+1]) ;
           }
           while(M--){
                scanf("%lld",&x) ;
                int id = find_id(x) ;
               // cout<<x<<"  "<<id<<endl ;
                if(id == -1){
                   printf("%lld
    ",P[N]*x) ;
                   continue ;
                }
                if(S[id] == x)
                   printf("%lld
    ",dp[id]) ;
                else
                   printf("%lld
    ",Min(dp[id],x*P[id-1])) ;
           }
        }
        return  0 ;
    }
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  • 原文地址:https://www.cnblogs.com/liyangtianmen/p/3440015.html
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