POJ 2976 Dropping tests (0/1分数规划)
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 4654 | Accepted: 1587 |
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 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
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
给定ai和bi,求100*sigma(ai*xi)/sigma(bi*xi)的最大值,xi=0或1,且只能有k个为0。
即可以去掉k个(ai,bi)对。
y=100*sigma(ai)/sigma(bi)
t(y)=100*sigma(ai)-y*sigma(bi)
求t(y0)=0;
当t(y)<0时,y太大,y0<y;
当t(y)>0时,y太小,y<y0;
所以可以二分y的值。
去哪k个更优了?
要使y更大就应该是t(y)>0这种情况更可能的出现,所以对100*ai-bi排序,去掉小的k个。使t(y)>0城里更有可能。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; const int N=1010; const double eps=1e-10; int n,k; double a[N],b[N],score[N]; int main(){ //freopen("input.txt","r",stdin); while(~scanf("%d%d",&n,&k)){ if(n==0 && k==0) break; for(int i=0;i<n;i++) scanf("%lf",&a[i]); for(int i=0;i<n;i++) scanf("%lf",&b[i]); double l=0,r=100,mid,sum; while(r-l>eps){ mid=(l+r)/2; for(int i=0;i<n;i++) score[i]=100*a[i]-mid*b[i]; sort(score,score+n); sum=0; for(int i=k;i<n;i++) sum+=score[i]; if(sum>0) l=mid; else r=mid; } printf("%.0f\n",l); } return 0; }