题目链接

A.怪盗-1412

题意：


题解：

#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

题意：

题解：



AC代码

/*
    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.序列卷积之和

题意：


题解：

AC代码

/*
    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.宝石装箱

题意：



题解：






















AC代码

/*
    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;
}