POJ3111 K Best,看讨论区说数据有点变态,精度要求较高,我就直接把循环写成了100次,6100ms过,(试了一下30,40都会wa,50是4000ms)
第一次在POJ上看到下面这种东西还是很好奇的,
一个题目可以接受多种正确答案,即有多组解的时候,题目就必须被Special Judge.
Special Judge程序使用输入数据和一些其他信息来判答你程序的输出,并将判答结果返回.
Case Time Limit: 2000MS | Special Judge |
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <sstream>
#include <fstream>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define in(n) scanf("%d",&(n))
#define in2(x1,x2) scanf("%d%d",&(x1),&(x2))
#define inll(n) scanf("%I64d",&(n))
#define inll2(x1,x2) scanf("%I64d%I64d",&(x1),&(x2))
#define inlld(n) scanf("%lld",&(n))
#define inlld2(x1,x2) scanf("%lld%lld",&(x1),&(x2))
#define inf(n) scanf("%f",&(n))
#define inf2(x1,x2) scanf("%f%f",&(x1),&(x2))
#define inlf(n) scanf("%lf",&(n))
#define inlf2(x1,x2) scanf("%lf%lf",&(x1),&(x2))
#define inc(str) scanf("%c",&(str))
#define ins(str) scanf("%s",(str))
#define out(x) printf("%d\n",(x))
#define out2(x1,x2) printf("%d %d\n",(x1),(x2))
#define outf(x) printf("%f\n",(x))
#define outlf(x) printf("%lf\n",(x))
#define outlf2(x1,x2) printf("%lf %lf\n",(x1),(x2));
#define outll(x) printf("%I64d\n",(x))
#define outlld(x) printf("%lld\n",(x))
#define outc(str) printf("%c\n",(str))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define mem(X,Y) memset(X,Y,sizeof(X));
typedef vector<int> vec;
typedef long long ll;
typedef pair<int,int> P;
const int dx[]={,,-,},dy[]={,,,-};
const double INF=0x3f3f3f3f;
const ll mod=1e9+;
ll powmod(ll a,ll b) {ll res=;a%=mod;for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}
const bool AC=true; int n,k;
struct point{
double w,v,c;
int num; //保存编号,便于输出答案
};
point p[];
bool cmp(point x,point y){
return x.c>y.c;
}
bool C(double x){
double sum=;
rep(i,,n){
p[i].c=p[i].v-p[i].w*x;
}
sort(p,p+n,cmp);
rep(i,,k){
sum+=p[i].c;
}
return sum>=;
}
int main(){
in2(n,k);
rep(i,,n){
inlf2(p[i].v,p[i].w);
p[i].num=i;
}
double lb=,ub=INF; rep(i,,){
double mid=(lb+ub)/;
if(C(mid)) lb=mid;
else ub=mid;
}
rep(i,,k){
printf("%d\n",p[i].num+);编号加一
}
return ;
}