【二分答案+01分數規劃搜尋】POJ - 2976 D - Dropping tests
D - Dropping tests POJ - 2976
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
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
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).
題目大意就 給定n個二元組(a,b),扔掉k個二元組,使得剩下的a元素之和與b元素之和的比率最大
題目求的是 max(∑a[i] * x[i] / (b[i] * x[i])) 其中a,b都是一一對應的。 x[i]取0,1 並且 ∑x[i] = n - k;
轉:那麼可以轉化一下。 令r = ∑a[i] * x[i] / (b[i] * x[i]) 則必然∑a[i] * x[i] - ∑b[i] * x[i] * r= 0;(條件1)
並且任意的 ∑a[i] * x[i] - ∑b[i] * x[i] * max(r) <= 0 (條件2,只有當∑a[i] * x[i] / (b[i] * x[i]) = max(r) 條件2中等號才成立)
所以
二分
const double eps=1e-7
while(r-l>=eps)
{
double mid=(l+r)/2.0;
if(ok(mid))
{
l=mid;
ans=mid;
}
else r=mid;
}
特別注意!!!
輸出不要用lf poj可能會WA printf("%f",ans);
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
double a[1050],b[1050],c[1050];
const double eps=1e-7;
int n,m;
bool cmp(const double &x,const double &y)
{
return x>y;
}
bool ok(double x)
{
for(int i=1;i<=n;i++)
c[i]=a[i]-b[i]*x;
sort(c+1,c+n+1,cmp);
double sum=0;
for(int i=1;i<=n-m;i++) sum+=c[i];
if(sum>0) return 1;
else return 0;
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
if(n==0&&m==0) break;
for(int i=1;i<=n;i++)
scanf("%lf",&a[i]);
for(int i=1;i<=n;i++)
scanf("%lf",&b[i]);
double l=0.0,r=1.0;
double ans=0;
while(r-l>=eps)
{
double mid=(l+r)/2.0;
if(ok(mid))
{
l=mid;
ans=mid;
}
else r=mid;
}
printf("%.0f\n",ans*100);
}
return 0;
}