A
#include<bits/stdc++.h>
#define debug(x) cout<<#x<<" is "<<x<<endl
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define DBG 0
const int N = 1e5 + 5;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const ll LLINF = (1LL<<60);
using namespace std;
const int mod = 998244353;
ll fast_pow(ll a,ll b){
ll ans = 1;
while(b){
if(b&1)ans = (ans * a)%mod;
a = (a * a)%mod;
b>>=1;
}
return (ans%mod);
}
typedef pair<int,int> pii;
int a[N],b[N];
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
#if DBG
freopen("input.txt","r",stdin);
#endif
int t;
cin>>t;
while(t--){
int n,x;
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,greater<int>());
bool f = true;
for(int i = 0; i< n;i++)
if(a[i] + b[i] > x)f = false;
printf("%s\n",f?"Yes":"No");
}
return 0;
}
B
#include<bits/stdc++.h>
#define debug(x) cout<<#x<<" is "<<x<<endl
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define DBG 0
const int N = 1e5 + 5;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const ll LLINF = (1LL<<60);
using namespace std;
const int mod = 998244353;
ll fast_pow(ll a,ll b){
ll ans = 1;
while(b){
if(b&1)ans = (ans * a)%mod;
a = (a * a)%mod;
b>>=1;
}
return (ans%mod);
}
typedef pair<int,int> pii;
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
#if DBG
freopen("input.txt","r",stdin);
#endif
int t;
cin>>t;
while(t--){
int a,b,c,d;
cin>>a>>b>>c>>d;
cout<<max(a+b,c+d)<<endl;
}
return 0;
}
C
#include<bits/stdc++.h>
#define debug(x) cout<<#x<<" is "<<x<<endl
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define DBG 0
const int N = 1e5 + 5;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const ll LLINF = (1LL<<60);
using namespace std;
const int mod = 998244353;
ll fast_pow(ll a,ll b){
ll ans = 1;
while(b){
if(b&1)ans = (ans * a);
a = (a * a);
b>>=1;
}
return ans;
}
typedef pair<int,int> pii;
int prime[N];
vector<int> pri;
void get(){
for(int i = 2;i <= 100000;i++){
if(!prime[i]){
pri.pb(i);
for(int j = i;j <= 100000;j += i)prime[j] = 1;
}
}
}
//int cnt1[N],cnt2[N];
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
#if DBG
freopen("input.txt","r",stdin);
#endif
get();
int t;
cin>>t;
while(t--){
ll p,q;
cin>>p>>q;
if(p % q == 0){
ll min_div= 1e18;
ll pp = p;
for(int i = 0;i < pri.size();i++){
if(q % pri[i] == 0){
ll cnt1= 0,cnt2= 0;
while(p % pri[i] == 0){
p/= pri[i];
cnt1++;
}
while(q % pri[i] ==0){
q/= pri[i];
cnt2++;
}
min_div = min(min_div,fast_pow(pri[i],cnt1 - cnt2 + 1));
}
}
if(q != 1){
ll cnt = 0;
while(p % q == 0)cnt++,p/=q;
min_div = min(min_div,fast_pow(q,cnt));
}
cout<<pp/min_div<<endl;
}else
cout<<p<<endl;
}
return 0;
}
D.考慮將2n個元素中較小的一半歸為集合L，較大的一半歸為集合R，
考慮某種劃分(p,q)，對于任意位置i, p_i,q_i 不可能都屬于L，
假定都屬于 L, 則 p_1p_i屬于L，q_iq_n 屬于L，則L有n+1個元素，矛盾，
同理p_i,q_i 不可能都屬于R，則p_i,q_i一個屬于L，一個屬于R，
任意劃分(p,q)對答案的貢獻都是sum?-sum(L)
#include<bits/stdc++.h>
#define debug(x) cout<<#x<<" is "<<x<<endl
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define DBG 0
const int N = 3e5 + 5;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const ll LLINF = (1LL<<60);
using namespace std;
const int mod = 998244353;
ll fast_pow(ll a,ll b){
ll ans = 1;
while(b){
if(b&1)ans = (ans * a)%mod;
a = (a * a)%mod;
b>>=1;
}
return (ans%mod);
}
typedef pair<int,int> pii;
ll a[N],b[N];
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
#if DBG
freopen("input.txt","r",stdin);
#endif
int n;
cin>>n;
for(int i = 1;i <= n;i++)
cin>>a[i];
for(int i = 1;i <= n;i++)
cin>>b[i];
sort(a + 1,a + 1 + n);
sort(b + 1,b + 1 + n,greater<int>());
ll ans = 0;
for(int i = 1;i <= n;i++)
ans += abs(a[i] - b[i]),ans %= mod;
for(int i = 2 * n;i > n;i--)
ans = (ans * i)%mod;
for(int i = n;i >= 1;i--)
ans = (ans * fast_pow(i,mod - 2))%mod;
cout<<ans<<endl;
return 0;
}
E
----------------待補-------------------
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