zoj 3278 8G Island(二分)

题意：给出n个元素的数组a和m个元素的数组b,求数组a中元素与数组b中元素相乘所得乘积的第k大值；

思路：暴力超时，采用二分；先将两数组从大到小排序，外层二分枚举答案，内层二分查找与当前枚举值相等的乘积所在的区间；

#include<cstdio>
#include<cstring>
#include<algorithm>
typedef long long ll;
#define inf 100000000010
using namespace std;
ll kth,n,m;
ll a[500010],b[500010];
int cmp(int a1,int b1)
{
    return a1>b1;
}
int main()
{
    int k,i,j;
    while(scanf("%lld%lld%lld",&n,&m,&kth)!=EOF)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(i=1;i<=n;i++) scanf("%lld",&a[i]);
        for(i=1;i<=m;i++) scanf("%lld",&b[i]);
        sort(a+1,a+n+1,cmp);
        sort(b+1,b+m+1,cmp);
        b[0]=inf;b[m+1]=-1;
        ll Min=a[n]*b[m];
        ll Max=a[1]*b[1];
        ll Low=Min;
        ll High=Max;
        ll ans=Max;
        while(Low<=High)
        {
            ll Mid=(Low+High)/2;
            ll key=Mid;
            ll s1=0,s2=0;
            for(i=1;i<=n;i++)
            {
                int low,high,res;
                ll val=a[i]*b[1];
                ll val2=a[i]*b[m];
                if(val2>key)
                {
                    s1+=m;
                    continue;
                }
                if(val<key)
                {
                    s2+=m;
                    continue;
                }
                low=1,high=m,res=0;
                while(low<=high)
                {
                    int mid=(low+high)/2;
                    val=a[i]*b[mid];
                    val2=a[i]*b[mid+1];
                    if(val>key&&val2<=key)
                    {
                        res=mid;
                        break;
                    }
                    else if(val2>key)
                        low=mid+1;
                    else
                        high=mid-1;
                }
                s1+=res;
                low=1,high=m,res=m+1;
                while(low<=high)
                {
                    int mid=(low+high)/2;
                    val=a[i]*b[mid];
                    val2=a[i]*b[mid-1];
                    if(val<key&&val2>=key)
                    {
                        res=mid;break;
                    }
                    else if(val2<key)
                    {
                        high=mid-1;
                    }
                    else
                        low=mid+1;
                }
                s2+=(m-res+1);
            }
            ll left=s1+1;
            ll right=n*m-s2;
            if(left<=kth&&kth<=right)
            {
                ans=key;break;
            }
            else if(kth<left)
            {
                Low=Mid+1;
            }
            else High=Mid-1;
        }
        printf("%lld
",ans);
    }
    return 0;
}