Zero Quantity Maximization

题目：

给定长度相等的数组a、b

c[i] = d*a[i] + b[i]

选择一个d使得c中0的个数尽可能多，问最多有多少0.

思路：

 考虑当a[i]为0时，只有b[i]也是0才满足条件

当a[i]不等于0时，计算b[i]/a[i]的值，用map记录并求解该键所对应的最大值

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<queue>
 6 #include<stack>
 7 #include <bitset>
 8 #include<set>
 9 #include<map>
10 #include<unordered_map>
11 #include<vector>
12 #include<cmath>
13 #include<string>
14 using namespace std;
15 typedef long long ll;
16 #define Maxn 200005
17 map<long double, ll>mp;
18 ll a[Maxn],b[Maxn];
19 ll n, res = 0, ans = 0;
20 int main() {
21     cin >> n;
22     for (int i = 0; i < n; i++) {
23         cin >> a[i];
24     }
25     for (int i = 0; i < n; i++) {
26         cin >> b[i];
27     //    c[i] = (-1) * b[i] / a[i];
28     }
29     for (int i = 0; i < n; i++) {
30         if (a[i] == 0) { //a[i]为0时，只有b[i]也是0,才能+1
31             if (b[i] == 0) { //直接跳过
32                 res++;
33             }
34             continue;
35         }
36         mp[(long double)b[i] / a[i]]++;
37         ans = max(ans, mp[(long double)b[i] / a[i]]);
38     }
39     
40     cout << ans + res << endl;
41     return 0;
42 }

 

posted @ 2020-07-20 22:32  吉吉的奥利奥  阅读(138)  评论(0)  编辑  收藏  举报