HDU 4791 & ZOJ 3726 Alice's Print Service (数学打表)

时间：2014-10-25 20:10:04 阅读：219 评论：0 收藏：0 [点我收藏+]

题目链接：

HDU:http://acm.hdu.edu.cn/showproblem.php?pid=4791

ZJU:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5072

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

HDU代码如下：

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

const int maxn=100017;
typedef __int64 LL;

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

int main()
{
    int T;
    int n, m;
    LL tt;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i = 0; i < n; i++)
        {
            scanf("%I64d%I64d",&s[i],&p[i]);
        }

        LL minn = INF;
        for(int i = n-1; i >= 0; i--)
        {
            minn = min(s[i]*p[i],minn);
            c[i] = minn;
        }
        for(int i = 0; i < m; i++)
        {
            scanf("%I64d",&tt);
            if(tt>=s[n-1])//最后
                printf("%I64d\n",tt*p[n-1]);
            else
            {
                int pos = upper_bound(s,s+n,tt)-s;
                LL ans = tt*p[pos-1];
                ans = min(ans,c[pos]);
                printf("%I64d\n",ans);
            }
        }
    }
    return 0;
}

ZJU代码如下：

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

const int maxn = 100017;
typedef long long LL;

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

int main()
{
    int T;
    int n, m;
    LL tt;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i = 0; i < n; i++)
        {
            scanf("%lld%lld",&s[i],&p[i]);
        }

        LL minn = INF;
        for(int i = n-1; i >= 0; i--)
        {
            minn = min(s[i]*p[i],minn);
            c[i] = minn;
        }
        for(int i = 0; i < m; i++)
        {
            scanf("%lld",&tt);
            if(tt>=s[n-1])//最后
                printf("%lld\n",tt*p[n-1]);
            else
            {
                int pos = upper_bound(s,s+n,tt)-s;
                LL ans = tt*p[pos-1];
                ans = min(ans,c[pos]);
                printf("%lld\n",ans);
            }
        }
    }
    return 0;
}

HDU 4791 & ZOJ 3726 Alice's Print Service (数学打表)

标签：zju hdu 数学

原文地址：http://blog.csdn.net/u012860063/article/details/40455549

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HDU 4791 & ZOJ 3726 Alice's Print Service (数学 打表)

HDU 4791 & ZOJ 3726 Alice's Print Service (数学打表)