题目链接：https://ac.nowcoder.com/acm/contest/11255/H
题目大意：定义运算符⊗：(x=prod p_i^{ a_i })，(y=prod p_i^{ b_i })，(xotimes y=prod p_i^{mid a_i-b_imid })，定义(b_i=sum_{1leq j,kleq n,jotimes k==i}^{}a_jk^c)，计算((b_1 xor b_2 xor ... xor b_n) mod 998244353)
题目思路：首先化简运算符(otimes)
[xotimes y=prod p_i^{mid a_i-b_imid } = prod p_i^{max(a_i,b_i)-min(a_i,b_i) } = frac{lcm(x,y)}{gcd(x,y)}=frac{xy}{gcd^2(x,y)}
]
这个(b_i)不好看，我们化成
[b_i=sum_{j=1}^{n}sum_{k=1}^{n}sum_{frac{jk}{gcd^2(j,k)}=i}^{}a_jk^c
]
枚举(g=gcd(j,k))
[b_i=sum_{j=1}^{n}sum_{k=1}^{n}sum_{g=1}sum_{frac{jk}{g^2}=i}a_jk^c
]
设(x=frac{j}{g},y=frac{k}{g})
[b_i=sum_{x=1}^{n}sum_{y=1}^{n}sum_{g=1}^{min(frac{n}{x},frac{n}{y})}sum_{xy=i}a_{jg}(yg)^c
]
后面和(g)没什么关系了，改变枚举顺序
[b_i=sum_{xy=i}y^csum_{g=1}^{min(frac{n}{x},frac{n}{y})}a_{jg}g^c
]
通过枚举(x,y,g)即可得出答案
AC代码：
#include <unordered_map>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <string>
#include <stack>
#include <deque>
#include <queue>
#include <cmath>
#include <map>
#include <set>
using namespace std;
typedef pair<int, int> PII;
typedef pair<double, int> PDI;
//typedef __int128 int128;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const int N = 1e6 + 10, M = 4e7 + 10;
const int base = 1e9;
const int P = 131;
const int mod = 998244353;
const double eps = 1e-9;
const double PI = acos(-1.0);
ll a[N], b[N];
ll dp[N], p[N];
int gcd(int a, int b)
{
return b == 0 ? a : gcd(b, a % b);
}
ll ksm(ll a, ll b)
{
ll res = 1 % mod;
while (b)
{
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
int main()
{
int n, c;
scanf("%d%d", &n, &c);
for (int i = 1; i <= n; ++i)
scanf("%lld", &a[i]);
for (int i = 1; i <= n; ++i)
p[i] = ksm(i, c);
for (int x = 1; x <= n; ++x)
{
for (int g = 1; g * x <= n; ++g)
dp[g] = (dp[g - 1] + p[g] * a[x * g]) % mod;
for (int y = 1; y * x <= n; ++y)
if (gcd(x, y) == 1)
b[x * y] = (b[x * y] + dp[min(n / x, n / y)] * p[y]) % mod;
}
ll ans = 0;
for (int i = 1; i <= n; ++i)
ans ^= b[i];
printf("%lld
", ans);
return 0;
}