Gym - 101194D(二分

最新推荐文章于 2021-04-23 10:41:57 发布

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本文介绍了一道算法题目——构造冰淇凌塔，通过给定的冰淇凌球数量及塔的高度要求，使用排序、二分查找及贪心算法来解决如何构造符合条件的冰淇凌塔的问题。

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构造冰淇凌塔需要下面一块是上面的两倍以上，给出n个冰淇凌球，要求塔高为k，问能构造出多少个冰淇凌塔，排序，二分可以构造出来的数量，然后贪心判断结果

//china no.1
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
typedef long long LL;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=3e5+10;
const int maxx=1e6+100;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

vector<pair<LL,LL > >v;
LL a[maxn];
int t,n,k;
int check(int x)
{
    v.clear();
    FOr(0,x,i)
        v.push_back(make_pair(0ll,0ll));//x个顶
    for(int i=1;i<=n;)
    {
        FOr(0,x,j)
        {
            W(a[i]<v[j].first*2&&i<=n)  i++;
            if(i>n) break;
            v[j].first=a[i];//更新顶
            v[j].second++;
            i++;
        }
    }
    FOr(0,x,i)
        if(v[i].second<k)
            return 0;
    return 1;
}

int main()
{
    scan_d(t);
    for(int cas=1;cas<=t;cas++)
    {
        scan_d(n);
        scan_d(k);
        FOR(1,n,i)
            scan_d(a[i]);
        sort(a+1,a+n+1);
        int l=0,r=n/k;
        int len=100;
        W(len--)
        {
            int mid=l+r>>1;
            if(check(mid)) l=mid+1;
            else r=mid-1;
        }
        printf("Case #%d: %d\n",cas,l-1);
    }
}