A.怪盜-1412
題意:
題解:
#include<bits/stdc++.h>
using namespace std;
int main(){
int _;
scanf("%d",&_);
while(_--){
long long n,m,k;
scanf("%lld%lld%lld",&n,&m,&k);
printf("%lld\n",(n/2)*m*k*(n-n/2));
}
}
B.Dis2
題意:
題解:
AC代碼
/*
Author:zzugzx
Lang:C++
Blog:blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int mod=1e9+7;
//const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=1e6+10;
const ll inf=0x3f3f3f3f;
const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
ll ans[maxn];
vector<int> g[maxn];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n;cin>>n;
for(int i=1,u,v;i<n;i++){
cin>>u>>v;
g[u].pb(v);
g[v].pb(u);
}
for(int u=1;u<=n;u++){
ll ans=0;
for(auto v:g[u]){
if(v==u)continue;
ans+=g[v].size()-1;
}
cout<<ans<<endl;
}
return 0;
}
C.序列卷積之和
題意:
題解:
AC代碼
/*
Author:zzugzx
Lang:C++
Blog:blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int mod=1e9+7;
//const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=1e6+10;
const ll inf=0x3f3f3f3f;
const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
ll Pow(ll a, ll b){
ll ans = 1;
while(b > 0){
if(b & 1){
ans = ans * a % mod;
}
a = a * a % mod;
b >>= 1;
}
return ans;
}
ll a[maxn],s1[maxn];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n;
cin>>n;
for(int i=1;i<=n;i++)
cin>>a[i];
for(int i=n;i;i--)
s1[i]=(s1[i+1]+a[i]*(n-i+1)%mod)%mod;
ll ans=0;
for(int i=1;i<=n;i++)
ans=(ans+i*a[i]%mod*s1[i]%mod)%mod;;
cout<<ans;
return 0;
}
D.寶石裝箱
題意:
題解:
AC代碼
/*
Author:zzugzx
Lang:C++
Blog:blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
//const int mod=1e9+7;
const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=1e6+10;
const ll inf=0x3f3f3f3f;
const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
ll fac[maxn],dp[maxn],a[maxn];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n;cin>>n;fac[0]=1;
for(int i=1,x;i<=n;i++){
cin>>x;
a[x]++;
fac[i]=fac[i-1]*i%mod;
}
dp[0]=1;
for(int i=1;i<=n;i++)
for(int j=i;j>=0;j--)
dp[j+1]=(dp[j+1]+dp[j]*a[i]%mod)%mod;
ll ans=fac[n],res=0;
for(int i=1,p=1;i<=n;i++,p=-p)
res=(res+fac[n-i]*dp[i]*p%mod+mod)%mod;
cout<<(ans-res+mod)%mod;
return 0;
}