题意:
给定n和k,求 1^k+2^k+3^k+...+(n−1)^k+n^k
n(1e9),k(<=3)
思路:分类k;
总结:公式!
k=1:(1+n)*(n)/2
k=2:
k=3:1^3+2^3+3^3+.+n^3=(1+2+3+.+n)^2
#include<bits/stdc++.h>
#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<sstream>
#include<cstdio>
#include<ctime>
#include<map>
#include<stack>
#include<string>
using namespace std;
#define sfi(i) scanf("%d",&i)
#define pri(i) printf("%d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define eps 1e-6
#define PI acos(-1)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define fl() printf("flag\n")
ll gcd(ll a,ll b){while(b^=a^=b^=a%=b);return a;}
const int maxn=3e6+9;
const int mod=998244353;
template <class T>
inline void sc(T &ret)
{
char c;
ret = 0;
while ((c = getchar()) < '0' || c > '9');
while (c >= '0' && c <= '9')
{
ret = ret * 10 + (c - '0'), c = getchar();
}
}
ll pow_mod(ll a, ll b, ll p){//a的b次方求余p
ll ret = 1;
while(b){
if(b & 1) ret = (ret * a) % p;
a = (a * a) % p;
b >>= 1;
}
return ret;
}
ll Fermat(ll a, ll p){//费马求a关于b的逆元
return pow_mod(a, p-2, p);
}
int main()
{
ll n,k;
while(cin>>n>>k)
{
n%=mod;
if(k==0)
{
cout<<n<<endl;
}
if(k==1)
{
ll ans=(1+n)*(n)/2;
cout<<ans%mod<<endl;
}
if(k==2)
{
ll ans=n*(n+1);
ans%=mod;
ans=ans*(2*n+1);
ans%=mod;
ans=ans*166374059;//6关于mod的逆元
cout<<ans%mod<<endl;
}
if(k==3)
{
ll ans=(1+n)*(n)/2;
ans%=mod;
ans=ans*ans%mod;
cout<<ans<<endl;
}
}
return 0;
}