zoj 3607 Lazier Salesgirl 暴力 前缀和
2015-04-19 21:29
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Lazier Salesgirl
Time Limit: 2 Seconds Memory Limit: 65536 KB
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?
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 containsn integers 1 ≤ ti ≤
100000. The customers are given in the non-decreasing order of ti.
For each test cases, output w and the corresponding average value of sold bread, with six decimal digits.
链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3607
题意:
某 在0时刻开店, 然后按顺序来了n个客人,分别出了一个价格p[i],每个人来的时刻为 t[i]。
然后 某 在w时间内没接待客人就会睡觉,如果间隔正好是w,不会睡。 问接待的客人 出价的平均值最大是多少,在满足该条件下 w最低是多少。
做法:
不断求前缀和 还有 到第i个人间隔最大值。 接待的人肯定是连续的 前多少个。 先假设只接待第一个人,然后假设接待前两个,再前三个。 类推。
如果是正好前i个人,那么w得 大于等于 ( 到第i个人间隔最大值) ,而且要严格小于(第 i和i+1 个人的间隔差)。 所以w就 取 ( 到第i个人间隔最大值),
如果这个值严格小于 (第 i和i+1 个人的间隔差)那么就成立,判断前i个人的平均价格是否更低。
Time Limit: 2 Seconds Memory Limit: 65536 KB
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 containsn 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
链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3607
题意:
某 在0时刻开店, 然后按顺序来了n个客人,分别出了一个价格p[i],每个人来的时刻为 t[i]。
然后 某 在w时间内没接待客人就会睡觉,如果间隔正好是w,不会睡。 问接待的客人 出价的平均值最大是多少,在满足该条件下 w最低是多少。
做法:
不断求前缀和 还有 到第i个人间隔最大值。 接待的人肯定是连续的 前多少个。 先假设只接待第一个人,然后假设接待前两个,再前三个。 类推。
如果是正好前i个人,那么w得 大于等于 ( 到第i个人间隔最大值) ,而且要严格小于(第 i和i+1 个人的间隔差)。 所以w就 取 ( 到第i个人间隔最大值),
如果这个值严格小于 (第 i和i+1 个人的间隔差)那么就成立,判断前i个人的平均价格是否更低。
#include <stdio.h> #include <stdlib.h> #include <string.h> #include <limits.h> #include <malloc.h> #include <ctype.h> #include <math.h> #include <string> #include <iostream> #include <algorithm> using namespace std; #include <stack> #include <queue> #include <vector> #include <deque> #include <set> #include <map> #define INF 999999999 #define eps 0.00001 #define LL __int64d #define pi acos(-1.0) struct node { int p, t; int sum; int big;//qian cha zui xiao int cha; }a[1047]; int cmp(node a,node b) { return a.sum>b.sum; } int main() { int t; scanf("%d",&t);//xiao w da pingjun while(t--) { int n, i, j; double sum = 0; scanf("%d",&n); a[0].sum=0; a[0].t=0; for(i = 1; i <= n; i++) { scanf("%d",&a[i].p); a[i].sum=a[i-1].sum+a[i].p; } for(i = 1; i <= n; i++) { scanf("%d",&a[i].t); if(i==1) a[i].big=a[i].t-a[i-1].t; else a[i].big=max(a[i-1].big,a[i].t-a[i-1].t); } for(i=1;i<=n;i++) { if(i==n) a[i].cha=99999999; else a[i].cha=a[i+1].t-a[i].t; } //sort(a+1,a+1+n,cmp); double bigavg=-1; double big; for(i=1;i<=n;i++) { if((1.0*a[i].sum/i)>bigavg) { if(a[i].cha>a[i].big) { big=a[i].big; bigavg=(1.0*a[i].sum/i); } } } printf("%.6lf %.6lf\n",big,bigavg); } return 0; }
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