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
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
题目大意就是一个买面包的小姑凉想偷懒，在卖到第i个人的时候，就会在一个之前维持的最大间隔时间内（如果在这个时间内没人来买面包的）她会一睡不醒。但又要满足能否去到最大平均值。
总体来说，这道题有两个条件：平均值最大，并且能一睡不醒。
1 #include<cstdio> 2 #include<string.h> 3 using namespace std; 4 double p[1010]; 5 double time[1010]; 6 double maxt[1010]; 7 double max(double a,double b) 8 { 9 return a>b?a:b; 10 } 11 int main() 12 { 13 int t,n; 14 double w; 15 double flag; 16 scanf("%d",&t); 17 while(t--) 18 { 19 scanf("%d",&n); 20 for(int i=1;i<=n;i++) 21 scanf("%lf",&p[i]); 22 for(int i=1;i<=n;i++) 23 scanf("%lf",&time[i]); 24 time[0]=0; 25 maxt[1]=time[1]-time[0]; 26 for(int i=2;i<=n;i++) 27 maxt[i]=max(time[i]-time[i-1],maxt[i-1]);//算出到第i个人时的前面的最大间隔时间 28 double maxn=0;//最大平均值 29 double anst=0;//间隔时间 30 double sum=0; 31 for(int i=1;i<=n;i++)//暴力枚举 32 { 33 w=maxt[i]; 34 sum+=p[i]; 35 if(i==n) 36 { 37 if(sum/i>maxn) 38 { 39 maxn=sum/i; 40 flag=w; 41 break; 42 } 43 } 44 if(sum/i>maxn&&w<time[i+1]-time[i])//要保证他卖给第i个人后能睡觉 45 { 46 maxn=sum/i; 47 flag=w; 48 } 49 50 } 51 printf("%.6lf %.6lf ",flag,maxn); 52 } 53 return 0; 54 }