20200727T3 【NOIP2017提高A组模拟8.10】计算几何

Description

 

3
4 5 3 
3 5 4 
2 
1 1 
3 3

0
3

 

Solution

然而并不用long long。

考场上就是直接排序然后判断 80分

可以用它的单调递增用二分优化就过了

 tip：作者发现自己菜得连二分都写不对了。调了一会。。

code

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<queue>
#include<vector>
#include<stack>
#include<set>
#include<deque>
#include<map>
using namespace std;

template <typename T> void read(T &x) {
    x = 0; int f = 1; char c;
    for (c = getchar(); c < '0' || c > '9'; c = getchar()) if (c == '-') f = -f;
    for (; c >= '0' && c <= '9'; c = getchar()) x = 10 * x + c - '0' ;
    x *= f;
}
template <typename T> void write(T x){
    if (x < 0) putchar('-'), x = -x;
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) { write(x); putchar('
'); }
template <typename T> void writesn(T x) { write(x); putchar(' '); }

#define ll long long
#define inf 100000
#define next net
#define P 9999991
#define N 100010
#define mid ((l+r)>>1)
#define lson (o<<1)
#define rson (o<<1|1)
#define R register
#define debug puts("zxt")

int n, m , aa, bb;
int x[N ], y[N ];
double k[N ], b[N ];
signed main()
{
    read(n);
    for(R int i = 1; i <= n; i++) read(x[i]);
    for(R int i = 1; i <= n; i++) read(y[i]);
    sort(x + 1, x + n + 1);
    sort(y + 1, y + n + 1);
    for(R int i = 1; i <= n;i++)
    {
        k[i] = - 1.0 * y[i] / x[i];
        b[i] = 1.0 * y[i];
    }
    read(m );
    while(m --)
    {
        read(aa); read(bb);
        int l = 0, r = n;
        while(l + 1 < r)
        {
            if(1.0 * aa * k[mid] + b[mid] <= (double)bb) l = mid;
            else r = mid - 1;
        }
        if(1.0 * aa * k[l + 1] + b[l + 1] <= (double)bb) writeln(l + 1);
        else writeln(l);
    }
    return 0;
}