【洛谷 4613】Olivander

题目描述

Harry Potter has damaged his magic wand in a fight with Lord Voldemort. He has decided to get a new wand in Olivander's wand shop. On the floor of the shop, he saw ​N wands and ​N wand boxes. The lengths of the wands are, respectively, X_1X1​ ,X_2X2​ ...​X_nXn​ , and the box sizes are Y_1Y1​,​Y_2Y2​ ...Y_nYn​ . A wand of length ​X can be placed in a box of size ​Y if ​X ≤ ​Y . Harry wants to know if he can place all the wands in boxes so that each box contains exactly one wand. Help him solve this difficult problem.

输入格式

The first line of input contains the positive integer ​N (1 ≤ ​N ≤ 100), the number from the task. The second line contains ​N positive integers ​X_iXi​ (1 ≤ ​X_iXi​ ≤ 10^9109​ ), the numbers from the task. The third line contains ​N positive integers ​X_iXi​ (1 ≤ X_iXi​ ≤ 10^9109​​ ), the numbers from the task.

输出格式

If Harry can place all the wands in boxes, output “DA” (Croatian for yes), otherwise output “NE” (Croatian for no).

输入输出样例

输入 #1复制

3
7 9 5
6 13 10

输出 #1复制

DA

输入 #2复制

4
5 3 3 5
10 2 10 10

输出 #2复制

NE

输入 #3复制

4
5 2 3 2
3 8 3 3

输出 #3复制

DA

说明/提示

In test cases worth 60% of total points, it will hold ​N ≤ 9.

Clarification of the first test case:

Harry can place the wands in boxes. For example, he can place the wand of length 5 in a box of size 6, wand of length 7 in a box of size 13, and wand of length 9 in a box of size 10.

Clarification of the second test case:

Harry can’t place the wands in boxes because the box of size 2 can’t fit any of the wands.

题解：一难配一简单哈哈。红题快乐。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<bits/stdc++.h>
using namespace std;
const int N=20;
int n,a[N],b[N];
bool ycll(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)
        scanf("%d",&b[i]);   
    sort(a+1,a+n+1);
    sort(b+1,b+n+1);
    for(int i=1;i<=n;i++)
        if(b[i]<a[i]) return 0;
    return 1;
}

int main(){
    freopen("4613.in","r",stdin);
    freopen("4613.out","w",stdout);
    int pp=ycll();
    if(pp==1) puts("DA");
    else puts("NE");
    return 0;
}