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二分学习

2024-09-08 06:38:10   214人阅读

题目描述

给定2个数组a、b以及一个整数x

判断是否存在a[i]+b[j]=x

输入

 第一行：两个整数n、m (1<n、m<10000)

第二行：n个整数，为数组a

第三行：m个整数，为数组b

第四行：一个整数k (1<k<20)，表示k次查询

接下来的k行：每行一个整数x

输出

 若数组a中存在一个a[i]，数组b中存在一个b[j]，使得a[i]+b[j]=x，则输出Yes

如不存在，则输出No

样例输入

2 3
1 2
3 4 5
4
1
4
5
8

样例输出

No
Yes
Yes
No

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 using namespace std;
 5 const int maxn = 1e5+7;
 6 int a[maxn], b[maxn];
 7 int query(int *arr, int l, int r, int key){
 8     while(l <= r){
 9         int mid = (l + r) >> 1;
10         if(arr[mid] == key) return mid;
11         if(arr[mid] < key) l = mid + 1;
12         else r = mid - 1;
13     }
14     return -1;
15 }
16 int main()
17 {
18     int n, m, k;
19     scanf("%d%d",&n,&m);
20     for(int i = 0; i < n; i++)scanf("%d",&a[i]);
21     for(int i = 0; i < m; i++)scanf("%d",&b[i]);
22     cin >> k;
23     sort(a, a + n);sort(b, b + m);
24     int q;
25     while(k--){
26         cin >> q;
27         int flag = 1;
28         for(int i = 0; i < n; i++){
29             int y = q - a[i];
30             //cout << y << endl;
31             if(y > 0){
32                 int pos = query(b, 0, m, y);
33                 if(pos != -1){
34                     cout << "Yes" << endl;
35                     flag = 0;
36                     break;
37                 }
38             }
39  
40         }
41         if(flag == 1){
42             cout << "No" << endl;
43         }
44     }
45     return 0;
46 }

 

 

670 D2

The term of this problem is the same as the previous one, the only exception — increased restrictions.

Input

The first line contains two positive integers n and k (1?≤?n?≤?100?000,?1?≤?k?≤?10^9) — the number of ingredients and the number of grams of the magic powder.

The second line contains the sequence a1,?a2,?...,?an (1?≤?ai?≤?10^9), where the i-th number is equal to the number of grams of the i-th ingredient, needed to bake one cookie.

The third line contains the sequence b1,?b2,?...,?bn (1?≤?bi?≤?10^9), where the i-th number is equal to the number of grams of the i-th ingredient, which Apollinaria has.

Output

Print the maximum number of cookies, which Apollinaria will be able to bake using the ingredients that she has and the magic powder.

Examples
input
1 1000000000
1
1000000000
output
2000000000
input
10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
1 1 1 1 1 1 1 1 1 1
output
0
input
3 1
2 1 4
11 3 16
output
4
input
4 3
4 3 5 6
11 12 14 20
output
3

 1 #include <iostream>
 2 #include <cstdio>
 3 #define ll long long
 4 using namespace std;
 5 const int maxn = 1e5+7;
 6 ll a[maxn], b[maxn];
 7 
 8 int main()
 9 {
10     ll n, k;
11     while(~scanf("%lld%lld",&n,&k)){
12         for(int i = 0; i < n; i++)scanf("%lld",&a[i]);
13         for(int i = 0; i < n; i++)scanf("%lld",&b[i]);
14         ll mid, sum, flag, l = 0, r = 2e9+10;
15         while(l <= r){
16             mid = (l + r) >> 1;
17             sum = 0;flag = 1;
18             for(int i = 0; i < n; i++){
19                 if(a[i]*mid > b[i]){
20                     sum += a[i]*mid - b[i];
21                 }
22                 if(sum > k){
23                     flag = 0;
24                 }
25             }
26             //cout << l << ‘ ‘ << r << endl;
27             if(flag) l = mid + 1;
28             else r = mid - 1;
29         }
30         printf("%lld\n",l - 1);
31     }
32 
33     return 0;
34 }

 

 







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