Dropping tests http://poj.org/problem?id=2976
Time Limit: 1000MS | Memory Limit: 65536K |
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
![](https://img.laitimes.com/img/_0nNw4CM6IyYiwiM6ICdiwiIml2ZuEzX2cTOy8CXzV2Zh1Wavw1Zy9mLq9Gcvw1LcpDc0RHaiojIsJye.gif)
.
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
Stanford Local 2005 题意:n种数,每种数有2个属性,a,b ,选出n-k种数,最大化∑A(n-k)/∑B(n-k) 01分数规划模板题 转换:令∑Ak/∑Bk=ans ,最大化ans 假设只有2个数 (A1+A2)/(B1+B2)=ans 转化:A1+A2=(B1+B2)*ans 去括号、移项:A1-B1*ans+A2-B2*ans=0 假设指定一个f,设ans为最终答案 若f<ans,那么式子>0 若f>ans,那么式子<0 所以二分ans,每次取出前k个最大(Ai-Bi*ans)判断是否>0 若>0,移动下界,否则,移动上界
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
int n,k;
double a[1001],b[1001],tmp[1001];
double l,r,mid,p,ans;
int main()
{
while(scanf("%d%d",&n,&k)!=EOF)
{
if(!n) break;
for(int i=1;i<=n;i++) scanf("%lf",&a[i]);
for(int i=1;i<=n;i++) scanf("%lf",&b[i]);
l=0;r=1;
while(fabs(l-r)>0.0001)
{
mid=(l+r)/2;p=0;
for(int i=1;i<=n;i++) tmp[i]=a[i]-mid*b[i];
sort(tmp+1,tmp+n+1);
for(int i=n;i>k;i--) p+=tmp[i];
if(p>0) l=mid;
else r=mid;
}
printf("%.0lf\n",l*100);
}
}
然而换了换二分姿势就错了,错误1:ans二分前ans要更新为0,防止不能二分
错误2:0.0001精度太小,卡到0.0000001就过了
至于为啥上面0.0001就能A,玄学
以后想着精度卡6、7位就好
错误代码:(就是用新的变量ans,当式子>0时,更新ans,最后输出ans)
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
int n,k;
double a[1001],b[1001],tmp[1001];
double l,r,mid,p,ans;
int main()
{
while(scanf("%d%d",&n,&k)!=EOF)
{
if(!n) break;
for(int i=1;i<=n;i++) scanf("%lf",&a[i]);
for(int i=1;i<=n;i++) scanf("%lf",&b[i]);
l=0;r=1;
while(fabs(l-r)>0.0001)
{
mid=(l+r)/2;p=0;
for(int i=1;i<=n;i++) tmp[i]=a[i]-mid*b[i];
sort(tmp+1,tmp+n+1);
for(int i=n;i>k;i--) p+=tmp[i];
if(p>=0) {ans=mid;l=mid+0.0001;}
else r=mid-0.0001;
}
printf("%.0lf\n",ans*100);
}
}
转载于:https://www.cnblogs.com/TheRoadToTheGold/p/6546981.html