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
.
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.
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.
3 1 5 0 2 5 1 6 4 2 1 2 7 9 5 6 7 9 0 0Sample Output
83 100Hint
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).
那我们怎么切入呢? 我们这样想
我需要sigama(ai)/sigama(bi)<=x(x就是我的最大值)
那么我是不是sigama(ai)-x*xigama(bi)<=0;
于是就有如下的代码了 注意最后是printf("%.0f);
或者你也可以强制int转换(int)(l*100+0.000000001);
欣赏代码吧
#include <cstdio> #include <cstring> #include <iostream> #include <cmath> #include <algorithm> using namespace std; #define dbg(x) cout<<#x<<" = "<< (x)<< endl const int MAX_N = 1024; int n,m; double x; struct node { int w; int s; bool operator < (const node& other) const{ return w-x*s>other.w-x*other.s; } }arr[MAX_N]; int check(double p){ double total_a = 0, total_b = 0; x = p; sort(arr+1,arr+1+n); for(int i = 1;i<=(n-m);++i){ total_a+=arr[i].w; total_b+=arr[i].s; } return total_a/total_b > p; } int main(){ while(~scanf("%d%d",&n,&m)&&(n||m)){ for(int i=1;i<=n;++i) scanf("%d",&arr[i].w); for(int i=1;i<=n;++i) scanf("%d",&arr[i].s); double l = 0,r=1; while(abs(r-l)>1e-4){ double mid =(l+r)/2; if(check(mid)) l=mid; else r=mid; } printf("%.0f\n",l*100); } return 0; }