POJ 1064 Cable master（二分查找+精度）（神坑题）

#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[4]={1,0,-1,0},dy[4]={0,1,0,-1};
const double INF=0x3f3f3f3f;
const ll mod=1e9+7;
ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
const bool AC=true;

double a[10005];
int n,k; //k为应该切割的绳子段数
int C(double x){ //计算切割成长度为x的绳子时能得到的绳子的段数(注意x为double)
    int ans=0;
    rep(i,0,n){
        ans+=floor(a[i]/x);
    }
    return ans;
}
int main(){
    in2(n,k);
    rep(i,0,n){
        inlf(a[i]);
    }
    double lb,mid,ub;
    lb=0.0,ub=INF;
    for(int i=0;i<100;i++){ //100次循环可以达到10^(-30)精度
        mid=(lb+ub)/2.0; //二分查找
        if(C(mid)<k) ub=mid; //切割的段数不够，要减小切割长度，此处不是C(mid)<=k
        //(因为要找到第一次等于k的值，此时若等于k,mid可能不是最大的值)
        else lb=mid; //切割段数满足，长度过小，要增大切割长度
    }
    printf("%.2f\n",floor(ub*100)/100);//一定要加floor函数,不加floor函数是四舍五入
    return 0;
}