POJ——T 2976 Dropping tests
http://poj.org/problem?id=2976
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 13861 | Accepted: 4855 |
Description
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 n integers indicating ai for all i. The third line contains npositive 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
Hint
To avoid ambiguities due to rounding errors, the judge tests have been constructed so that all answers are at least 0.001 away from a decision boundary (i.e., you can assume that the average is never 83.4997).
Source
1 #include <algorithm> 2 #include <cstdio> 3 4 inline void read(int &x) 5 { 6 x=0; register char ch=getchar(); 7 for(; ch>'9'||ch<'0'; ) ch=getchar(); 8 for(; ch>='0'&&ch<='9'; ch=getchar()) x=x*10+ch-'0'; 9 } 10 const double eps(1e-9); 11 const int N(1005); 12 double a[N],b[N]; 13 int n,k; 14 15 double l,r,mid,ans,tmp[N]; 16 inline bool check(double x) 17 { 18 double sum=0.0; 19 for(int i=1; i<=n; ++i) 20 tmp[i]=1.0*b[i]*x-100.0*a[i]; 21 std::sort(tmp+1,tmp+n+1); 22 for(int i=1; i<=n-k; ++i) sum+=tmp[i]; 23 return sum<0; 24 } 25 26 int Presist() 27 { 28 for(; 1; ) 29 { 30 read(n),read(k); if(!n&&!k) break; 31 for(int i=1; i<=n; ++i) scanf("%lf",&a[i]); 32 for(int i=1; i<=n; ++i) scanf("%lf",&b[i]); 33 for(l=0,r=100.0; r-l>eps; ) 34 { 35 mid=(l+r)/2.0; 36 if(check(mid)) 37 l=mid; 38 else r=mid; 39 } 40 printf("%.0lf\n",l); 41 } 42 return 0; 43 } 44 45 int Aptal=Presist(); 46 int main(int argc,char**argv){;}