POJ 2976 Dropping tests

Dropping tests

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on  PKU. Original ID: 2976
64-bit integer IO format: %lld      Java class name: Main

 

In a certain course, you take n tests. If you get ai out of bi questions correct on test i, your cumulative average is defined to be

.

Given your test scores and a positive integer k, determine how high you can make your cumulative average if you are allowed to drop any k of your test scores.

Suppose you take 3 tests with scores of 5/5, 0/1, and 2/6. Without dropping any tests, your cumulative average is . However, if you drop the third test, your cumulative average becomes .

 

Input

The input test file will contain multiple test cases, each containing exactly three lines. The first line contains two integers, 1 ≤ n ≤ 1000 and 0 ≤ k < n. The second line contains nintegers indicating ai  for all i. The third line contains n positive integers indicating bi for all i. It is guaranteed that 0 ≤ ai ≤ bi ≤ 1, 000, 000, 000. The end-of-file is marked by a test case with n = k = 0 and should not be processed.

Output

For each test case, write a single line with the highest cumulative average possible after dropping k of the given test scores. The average should be rounded to the nearest integer.

 

Sample Input

3 1

5 0 2

5 1 6

4 2

1 2 7 9

5 6 7 9

0 0

Sample Output

83

100

Source

Stanford Local 2005

 

解题：01分数规划。Dinkelbach

 

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <climits>

 7 #include <vector>

 8 #include <queue>

 9 #include <cstdlib>

10 #include <string>

11 #include <set>

12 #include <stack>

13 #define LL long long

14 #define pii pair<int,int>

15 #define INF 0x3f3f3f3f

16 using namespace std;

17 const double exps = 1e-6;

18 struct node{

19     double c,a,b;

20 };

21 node p[1010];

22 int n,k;

23 bool cmp(const node &x,const node &y){

24     return x.c < y.c;

25 }

26 double test(){

27     double ans,x,y,tmp = 0;

28     while(true){

29         ans = tmp;

30         for(int i = 0; i < n; i++)

31             p[i].c = p[i].a - ans*p[i].b;

32         sort(p,p+n,cmp);

33         x = y = 0;

34         for(int i = k; i < n; i++){

35             x += p[i].a;

36             y += p[i].b;

37         }

38         tmp = x*1.0/y;

39         if(fabs(tmp - ans) < exps) return ans;

40     }

41 }

42 int main() {

43     while(scanf("%d %d",&n,&k),n||k){

44         for(int i = 0; i < n; i++)

45             scanf("%lf",&p[i].a);

46         for(int i = 0; i < n; i++)

47             scanf("%lf",&p[i].b);

48         printf("%.0f\n",test()*100);

49     }

50     return 0;

51 }

View Code