ZOJ 3607 Lazier Salesgirl (贪心,模拟)
2017-04-15 13:41
471 查看
Description
Kochiya Sanae is a lazy girl who makes and sells bread. She is an expert at bread making and selling. She can sell the i-th customer a piece of bread for price pi. But she is so lazy that she will fall asleep if no customer
comes to buy bread for more than w minutes. When she is sleeping, the customer coming to buy bread will leave immediately. It's known that she starts to sell bread now and the i-th customer come after ti minutes.
What is the minimum possible value of w that maximizes the average value of the bread sold?
Input
There are multiple test cases. The first line of input is an integer T ≈ 200 indicating the number of test cases.
The first line of each test case contains an integer 1 ≤ n ≤ 1000 indicating the number of customers. The second line contains n integers 1 ≤ pi ≤ 10000. The third line contains n integers 1 ≤ ti ≤
100000. The customers are given in the non-decreasing order of ti.
Output
For each test cases, output w and the corresponding average value of sold bread, with six decimal digits.
Sample Input
Sample Output
题意:一个小姑娘烤面包卖,先给出n个客户,他们会在t[i]时间内买价值p[i]的面包,但是这个姑娘有个最大休息时间,如果每隔w时间,没有人来买面包,他就不卖了,求出使得每次平均面包价值(卖出面包总价值/卖的次数)最大的情况下,最小的间隔时间。
思路:先求出要卖出第i个面包,最小时间间隔wi[i]。在O(N)复杂度下,遍历每个顾客,当遇到一个顾客正在买第i个面包时,再想要等到第i+1个顾客需要延迟minw时,就可以计算一下当前的平均价值,并且与之前的maxave比较,取最大值,并记录minw。
注意:小女孩卖面包不是想停就能停的,需要在w间隔内没人来,才能停下,如果当前取得时间间隔minw>t[i + 1] - t[i],这种情况下,哪怕达到的值很大,也不能停下。
代码如下:#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
int p[1005];
int t[1005];
int wi[1005];
int main(){
int cases, n, i, j, pre;
double ave, maxave, minw, sum;
cin>>cases;
while(cases--){
cin>>n;
pre = 0;
sum = 0;
maxave = 0;
minw = 0;
for(i = 0; i < n; i++){
cin>>p[i];
}
for(i = 0; i < n; i++){//计算得要卖到当前的面包,最小间隔
cin>>t[i];
wi[i] = max(t[i] - pre, wi[i - 1]);
pre = t[i];
}
for(i = 0; i < n; i++){
sum += p[i];
if(i == n - 1 || wi[i + 1] > wi[i]){//如果要卖出下一个面包,需要更大的时间间隔或者已经到了最后一个,则计算一下当前的平均值
if(sum / (i + 1) > maxave){
maxave = sum / (i + 1);
minw = wi[i];
}
}
}
printf("%.6lf %.6lf\n", minw, maxave);
}
return 0;
}
Kochiya Sanae is a lazy girl who makes and sells bread. She is an expert at bread making and selling. She can sell the i-th customer a piece of bread for price pi. But she is so lazy that she will fall asleep if no customer
comes to buy bread for more than w minutes. When she is sleeping, the customer coming to buy bread will leave immediately. It's known that she starts to sell bread now and the i-th customer come after ti minutes.
What is the minimum possible value of w that maximizes the average value of the bread sold?
Input
There are multiple test cases. The first line of input is an integer T ≈ 200 indicating the number of test cases.
The first line of each test case contains an integer 1 ≤ n ≤ 1000 indicating the number of customers. The second line contains n integers 1 ≤ pi ≤ 10000. The third line contains n integers 1 ≤ ti ≤
100000. The customers are given in the non-decreasing order of ti.
Output
For each test cases, output w and the corresponding average value of sold bread, with six decimal digits.
Sample Input
2 4 1 2 3 4 1 3 6 10 4 4 3 2 1 1 3 6 10
Sample Output
4.000000 2.500000 1.000000 4.000000
题意:一个小姑娘烤面包卖,先给出n个客户,他们会在t[i]时间内买价值p[i]的面包,但是这个姑娘有个最大休息时间,如果每隔w时间,没有人来买面包,他就不卖了,求出使得每次平均面包价值(卖出面包总价值/卖的次数)最大的情况下,最小的间隔时间。
思路:先求出要卖出第i个面包,最小时间间隔wi[i]。在O(N)复杂度下,遍历每个顾客,当遇到一个顾客正在买第i个面包时,再想要等到第i+1个顾客需要延迟minw时,就可以计算一下当前的平均价值,并且与之前的maxave比较,取最大值,并记录minw。
注意:小女孩卖面包不是想停就能停的,需要在w间隔内没人来,才能停下,如果当前取得时间间隔minw>t[i + 1] - t[i],这种情况下,哪怕达到的值很大,也不能停下。
代码如下:#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
int p[1005];
int t[1005];
int wi[1005];
int main(){
int cases, n, i, j, pre;
double ave, maxave, minw, sum;
cin>>cases;
while(cases--){
cin>>n;
pre = 0;
sum = 0;
maxave = 0;
minw = 0;
for(i = 0; i < n; i++){
cin>>p[i];
}
for(i = 0; i < n; i++){//计算得要卖到当前的面包,最小间隔
cin>>t[i];
wi[i] = max(t[i] - pre, wi[i - 1]);
pre = t[i];
}
for(i = 0; i < n; i++){
sum += p[i];
if(i == n - 1 || wi[i + 1] > wi[i]){//如果要卖出下一个面包,需要更大的时间间隔或者已经到了最后一个,则计算一下当前的平均值
if(sum / (i + 1) > maxave){
maxave = sum / (i + 1);
minw = wi[i];
}
}
}
printf("%.6lf %.6lf\n", minw, maxave);
}
return 0;
}
相关文章推荐
- ZOJ-3607-Lazier Salesgirl【模拟】【贪心】【9th浙江省赛】
- ZOJ - 3607 Lazier Salesgirl(模拟)
- ZOJ 3607-Lazier Salesgirl-贪心
- ZOJ 3607 Lazier Salesgirl 贪心
- ZOJ 3607 Lazier Salesgirl 贪心
- zoj 3607 Lazier Salesgirl 贪心
- ZOJ 3607 Lazier Salesgirl (贪心)
- ZOJ 3607 Lazier Salesgirl (贪心)
- ZOJ 3607 Lazier Salesgirl(贪心啊 )
- ZOJ _3607_Lazier Salesgirl(贪心)
- ZOJ 3607 Lazier Salesgirl【贪心】
- ZOJ 3607 Lazier Salesgirl
- 【有点坑】ZOJ 3607 Lazier Salesgirl
- ZOJ Problem Set - 3607 Lazier Salesgirl
- zoj 3607 Lazier Salesgirl
- zoj 3607 Lazier Salesgirl
- zoj 3607 Lazier Salesgirl 暴力 前缀和
- ZOJ 3607 Lazier Salesgirl
- ZOJ 3607 Lazier Salesgirl(greed)
- ZOJ 3607 Lazier Salesgirl