A 水题
代码语言:javascript复制#include<bits/stdc .h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
int n,m;
cin>>n>>m;
if(n % m==0) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
}
B 水题
代码语言:javascript复制#include<bits/stdc .h>
using namespace std;
int a[105];
int main(){
int t;
cin>>t;
while(t--){
int n;
cin>>n;
for(int i=0;i<n;i ){
cin>>a[i];
}
sort(a,a n);
for(int i=n-1;i>=1;i--) cout<<a[i]<<" ";
cout<<a[0]<<endl;
}
return 0;
}
C水题
代码语言:javascript复制#include<bits/stdc .h>
using namespace std;
typedef long long LL;
int T;
int n,k;
int a[70];
LL a;
int main()
{
cin>>T;
while(T--)
{
memset(a,0,sizeof(a))
cin>>n>>k;
bool flag=0;
for(int i=0;i<n;i )
{
cin>>a;
LL tmp=a;
int j=0;
while(tmp)
{
a[j] =(tmp%k);
if(a[j]>1){flag=1;break;}
j ;
tmp=tmp/k;
}
}
if(!flag)puts("YES");
else puts("NO");
}
return 0;
}
D. 组合问题
代码语言:javascript复制#include<bits/stdc .h>
using namespace std;
#define ll long long
#define mod 998244353
const ll maxn=2e5 10;
ll n,m,t;
ll ksm(ll a,ll b)
{
if(b<0)
return 0;
ll ans=1;
while(b)
{
if(b&1)
ans=ans*a%mod;
a=a*a%mod;
b>>=1;
}
return ans;
}
ll exgcd(ll a,ll b,ll &x,ll &y){
if(!b){
x=1;y=0;
return a;
}
ll d=exgcd(b,a%b,x,y);
ll tmp=x;
x=y;
y=tmp-a/b*y;
return d;
}
ll inv(ll a,ll m){
ll x,y;
ll d=exgcd(a,m,x,y);
if(d==1){
return (x%m m)%m;
}
return -1;
}
int main(){
cin>>n>>m;
ll nn=n-1;
ll sum=0;
ll a=1,b=1;int f=1;
for(ll i=m;i>=m-nn 1 ;i--){
a*=i;a%=mod;
b*=(m-i 1);b%=mod;
}
sum=a*inv(b,mod)%mod;
ll k=(n-2)*ksm(2,n-3)%mod;
sum*=k;
sum%=mod;
cout<<sum<<endl;
return 0;
}
E 大佬题解 表示不会,dp没想到
代码语言:javascript复制#include<bits/stdc .h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
typedef long long LL;
inline int read()
{
int x=0,f=1;char c=getchar();
while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
while(isdigit(c)){x=x*10 c-'0';c=getchar();}
return x*f;
}
const int maxn=510;
int n,a[maxn],dp[maxn][maxn],vis[maxn][maxn],f[maxn];
int DP(int L,int R)
{
if(vis[L][R])return dp[L][R];
vis[L][R]=1;
int &res=dp[L][R];
if(R<L)return -1;
if(R==L)return res=a[L];
for(int i=L;i<R;i )
if(DP(L,i)==DP(i 1,R) && DP(L,i)!=-1)return res=DP(L,i) 1;
return res;
}
int main()
{
mem(dp,-1);
n=read();
for(int i=1;i<=n;i )a[i]=read();
DP(1,n);
mem(f,42);
f[0]=0;
for(int i=1;i<=n;i )
for(int j=1;j<=i;j )
if(dp[j][i]!=-1)f[i]=min(f[i],f[j-1] 1);
printf("%dn",f[n]);
return 0;
}