Avito Cool Challenge 2018

我好菜鸭

A.除了二，其他都可以减n-1，所以除了二都能减到一

#include<cstdio>
int main()
{
    int n;scanf(\"%d\",&n);
    if(n!=2)
	printf(\"1\\n\");
	else
	printf(\"2\\n\");
    return 0;
}

B.对于数字相同的，颜色可能相同，把一个数字看作一个颜色则该颜色的个数为n-arr[i]

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int arr[100010],g[100010];
int color[100010];
bool cmp(int a,int b)
{
	return arr[a]<arr[b];
}
int main()
{
	int n;cin>>n;
	for(int i=0;i<n;i++)
	{
		cin>>arr[i];
	}
	for(int i=0;i<=n;i++)
		g[i]=i;
	sort(g,g+n,cmp);
	bool ok=1;
	int j=1;
	for(int i=0;i<n;)
	{
		int u=arr[g[i]];
		int d = n - u;
		for(int k=0;k<d;k++)
			if(arr[g[i+k]]!=u)
				ok=0;	
			else
				color[g[i+k]]=j;
		i=i+d;
		j++;
	}
	if(ok)
	{
		cout<<\"Possible\"<<endl;
		for(int i=0;i<n;i++)
			cout<<color[i]<<\" \";
			cout<<endl;
	}		
	else
	cout<<\"Impossible\"<<endl;
	return 0;
}

C.推出公式C(n-1,k)*m*(m-1)^k

#include<cstdio>
#define ll long long
const int N = 2000 + 5;
const ll MOD = 998244353;
int comb[N][N];
void init(){
    for(int i = 0; i < N; i ++){
        comb[i][0] = comb[i][i] = 1;
        for(int j = 1; j < i; j ++){
            comb[i][j] = comb[i-1][j] + comb[i-1][j-1];
            comb[i][j] %= MOD;
        }
    }
}
ll qpow(ll a,ll b)
{
	ll ret=1;
	while(b)
	{
		if(b&1)ret=(ret*a)%MOD;
		a=(a*a)%MOD;
		b>>=1;
	}
	return ret%MOD;
}
int main(){
    init();
    int n,m,k;
    scanf(\"%d%d%d\",&n,&m,&k);
    printf(\"%I64d\\n\",((comb[n-1][k]%MOD*m%MOD*qpow(m-1,k)%MOD)%+MOD)%MOD);  
	return 0;
}