【POJ】2976 Dropping tests（二分）
【POJ】2976 Dropping tests（二分）
【题目链接】http://poj.org/problem?id=2976
题目内容
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).
题目大意
给你两个数组a[],b[]，分别代表每次测试得到的分和总分，可以除掉其中的k对，要使sum(a[i])/sum(b[i])*100，也就是剩下的占分比例最大。
解题思路
一开始想用贪心，但是找不出合适的规律尤其是数据中包括0，所以放弃了233.
直接二分删除后的sum(a)/sum(b),减去对这个答案贡献最小的几对，看能不能取到这个答案，在我们要找的最终答案的左侧都是可以取到的，右侧都不能。注意最后输出时四舍五入。
AC代码
#includeint main()
{
int n,k;
while(scanf("%d%d",&n,&k)!=EOF&&!(n==0&&k==0))
{
for(int i=0;iscanf("%lld",&p[i].a);
}
for(int i=0;iscanf("%lld",&p[i].b);
}
double left=0.0,right=1.1,mid,ans,x,y;
while(fabs(right-left)>1e-8)
{
mid=(right+left)/2;
for(int i=0;idouble)p[i].b*mid;//a中超出b*mid的那部分就是这组数据对最终结果能造成的影响，正数是提高，负数是减少。
}
sort(p,p+n,cmp);
ans=0;
for(int i=k;iif(ans>=0)
left=mid;
else right=mid;
}
printf("%.0f\n",100*left);//浮点数输出会自动四舍五入
}
return 0;
}