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 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
/*裸的01分數規劃問題.令∑a[i]/∑b[i]=ans. 則∑a[i]-∑b[i]*ans=0. 二分一個ans.然后用a[i]-b[i]*ans取前k大檢驗.只能去感性的認識orz...并不會證明.. */#include<iostream>#include<cstdio>#include<algorithm>#define eps 1e-7#define MAXN 1001using namespace std;double ans,a[MAXN],b[MAXN],sum,tmp[MAXN];int n,m,k;bool check(double x){ double tot=0; for(int i=1;i<=n;i++) tmp[i]=a[i]-x*b[i]; sort(tmp+1,tmp+n+1,greater<double>()); for(int i=1;i<=n-k;i++) tot+=tmp[i]; if(tot>=0) return true; else return false;}void slove(){ double l=0,r=1e4,mid; while(l<=r) { mid=(l+r)/2.0; if(check(mid)) l=mid+eps,ans=mid; else r=mid-eps; } printf("%.0f/n",ans*100); return ;}int main(){ while(scanf("%d%d",&n,&k)) { if(!n&&!k) break; sum=ans=0; for(int i=1;i<=n;i++) scanf("%lf",&a[i]); for(int i=1;i<=n;i++) scanf("%lf",&b[i]); slove(); } return 0;}/*發現這題牛頓迭代可做吖.網上的題解都是二分01規劃的.我就寫個牛頓迭代的吧orz(雖然二分的寫過).先選一個估計值s0.我們能保證這個答案是單調的.假設上次迭代的ans為s1,則存在n-k個元素使s1=∑(ai/bi),變形可得到∑ai-s2*∑bi=0,令ans[i]=a[i]-b[i]*s0.取前n-k大統計一個答案.可知必存在n-k個元素使∑ansi=∑ai-s1*∑bi=0,所以當我們按ans排序并取前n-k個元素作為求其∑ans時,∑ansi顯然是>=0的,然后s1=(∑ai-∑ansi)/∑bi)<=(∑ai/∑bi)=s2(i<=n-k).即此迭代過程是收斂的,當等號成立時,s即為答案.有些地方還是有點想不通畢竟弱吖orz. */#include<cstdio>#include<algorithm>#include<cmath>#define MAXN 1001#define eps 1e-7using namespace std;double ans,sum,tmp[MAXN];int n,m,k;struct data{double a,b,ans;}s[MAXN];bool cmp(const data &x,const data &y){ return x.ans>y.ans;}void slove(){ double suma=0,sumb=0,s0=0,s1=0; for(int i=1;i<=k;i++) suma+=s[i].a,sumb+=s[i].b; s0=suma/sumb; while(abs(s0-s1)>eps) { s1=s0;suma=sumb=0; for(int i=1;i<=n;i++) s[i].ans=s[i].a-s[i].b*s0; sort(s+1,s+n+1,cmp); for(int i=1;i<=k;i++) suma+=s[i].a,sumb+=s[i].b; s0=suma/sumb; } printf("%.0f/n",s0*100); return ;}int main(){ while(~scanf("%d%d",&n,&k)) { if(!n&&!k) break; sum=ans=0;k=n-k; for(int i=1;i<=n;i++) scanf("%lf",&s[i].a); for(int i=1;i<=n;i++) scanf("%lf",&s[i].b); slove(); } return 0;}新聞熱點
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