ZOJ 3607Lazier Salesgirl

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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 p_i. 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 t_i 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 ≤ p_i ≤ 10000. The third line contains n integers 1 ≤ t_i ≤ 100000. The customers are given in the non-decreasing order of t_i.

Output

For each test cases, output w and the corresponding average value of sold bread, with six decimal digits.

Sample Input

241 2 3 41 3 6 1044 3 2 11 3 6 10

Sample Output

4.000000 2.5000001.000000 4.000000
简单递推
#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<set>#include<ctime>#include<vector>#include<cmath>#include<algorithm>#include<map>#define ll long longusing namespace std;const int maxn = 1e5 + 10;int T, n, p[maxn], t[maxn];int main(){scanf("%d", &T);while (T--){scanf("%d", &n);for (int i = 1; i <= n; i++) scanf("%d", &p[i]);for (int i = 1; i <= n; i++) scanf("%d", &t[i]);int now = 0, w = 0, sum = t[0] = 0;double ans = 0;for (int i = 1; i <= n; ){while (i <= n&&now >= t[i] - t[i - 1]) sum += p[i++];if (i&&1.0*sum / (i - 1) > ans){ans = 1.0*sum / (i - 1);w = now;}now = t[i] - t[i - 1];}printf("%.6lf %.6lf\n", 1.0*w, 1.0*ans);}return 0;}

温故而知新#include<map>#include<cmath>    #include<queue>    #include<vector>#include<cstdio>    #include<cstring>    #include<algorithm>    using namespace std;#define ms(x,y) memset(x,y,sizeof(x))    #define rep(i,j,k) for(int i=j;i<=k;i++)    #define per(i,j,k) for(int i=j;i>=k;i--)    #define loop(i,j,k) for (int i=j;i!=-1;i=k[i])    #define inone(x) scanf("%d",&x)    #define intwo(x,y) scanf("%d%d",&x,&y)    #define inthr(x,y,z) scanf("%d%d%d",&x,&y,&z)    typedef long long LL;const int low(int x) { return x&-x; }const int INF = 0x7FFFFFFF;const int mod = 1e9 + 7;const int N = 1e3 + 10;int T, n, p[N], t[N];int main(){for (inone(T); T--;){inone(n); rep(i, 1, n) inone(p[i]);rep(i, 1, n) inone(t[i]);t[0] = 0;double w = 0, a = 0, g = 0, s = 0;for (int i = 1; i <= n; ){g = t[i] - t[i - 1];while (i <= n && t[i] - t[i - 1] <= g) s += p[i++];if (s / (i - 1) > a) a = s / (i - 1), w = g;}printf("%.6lf %.6lf\n", w, a);}return 0;}

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