题意: 给出两个数组a和b,将它们重新排序,如果满足每个a[i]+b[i] <= x则输出Yes,否则输出No。
思维: 不难想到这肯定得最大的和最小的相匹配,所以一个数组升序一个数组降序再判断一下即可。
代码实现:
#include<bits/stdc++.h>
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {
{
1, 0}, {
-1, 0}, {
0, 1}, {
0, -1}};
using namespace std;
const int inf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll mod = 1e9 + 7;
const int N = 2e5 + 5;
int t, n, x;
int a[N], b[N];
int cmp(int x, int y){
return x > y;
}
signed main()
{
IOS;
cin >> t;
while(t --){
cin >> n >> x;
for(int i = 0; i < n; i++) cin >> a[i];
for(int i = 0; i < n; i++) cin >> b[i];
sort(a, a+n); sort(b, b+n, cmp);
int flag = 1;
for(int i = 0; i < n; i ++)
if(a[i]+b[i]>x) {
flag = 0; break;}
cout << (flag ? "Yes":"No") << endl;
}
return 0;
}