[poj P2976] Dropping tests
[poj P2976] Dropping tests
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
.
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 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).
Source
01分数规划入门题。
这个不等式会经常看到:sigma(ai)/sigma(bi)>=(或<=)k
对于这题来说就是,找出最大的k,使得100*sigma(ai)/sigma(bi)>=k。
我们尝试化简上式。
因为bi>=0,所以sigma(bi)>=0,100*sigma(ai)>=k*sigma(bi)
100*sigma(ai)-k*sigma(bi)>=0
sigma(100*ai-k*bi)>=0
那么,我们只要先枚举一个k,将100*ai-k*bi作为关键字排序,再选出前n-k大的,判断一下sum是否非负就行了。
然后我们发现,上式满足单调性,枚举可以改为用二分。所以总复杂度是O(nlog分数)。
然后这一题我先开大100倍,然后最后再缩小100倍,但是这也会有精度误差。
总之,开大的倍数越大,误差越小,但效率越低,但也低不到哪里去qwq
code:
#include<cstdio> #include<cstring> #include<algorithm> #define LL long long #define M ((L)+(R)>>1) using namespace std; ,lim=; int n,k,L,R,ans,final; LL sum; struct ob {LL a,b,c;}o[N]; bool cmp(ob x,ob y) {return x.c>y.c;} bool jug(LL lv) { ; i<=n; i++) o[i].c=o[i].a*lim-lv*o[i].b; sort(o+,o++n,cmp); sum=; ; i<=n-k; i++) sum+=o[i].c; ; } int main() { while (scanf("%d%d",&n,&k)!=EOF,n|k) { ; i<=n; i++) scanf("%lld",&o[i].a); ; i<=n; i++) scanf("%lld",&o[i].b); ,R=lim; L<=R; ) ; ; final=ans/; ans%=; ) final+=; printf("%d\n",final); } ; }
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