牛客 Wannafly挑战赛27 D 绿魔法师

传送门

\(\color{green}{solution}\)

分析下,在\(1e5+1\)内,一个数的约数个数最多为\(2^{6}\)个,所以我们可以考虑枚举约数
复杂度\(O(N^{2^{6 \times 2}})\),实际上远远不到

#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;

vector <int> g[maxn];

int _pow(int x, int n, int P) {
    int ret = 1;
    for ( ; n; n >>= 1, x = 1LL * x * x % P)
        if( n & 1) ret = 1LL * ret * x % P;
    return ret;
}

int val[maxn], ins[maxn];

inline void qry(int val, int opt) {
    for ( register int i = 0; i < g[val].size(); ++ i) {
        ins[g[val][i]] += opt;
    }
}

int solve(int x, int K, int P) {
    memset(ins, 0, sizeof(ins));
    int ret = _pow(x, K, P);
    for ( register int i = 0; i < g[x].size(); ++ i) {
        if( val[g[x][i]]-ins[g[x][i]]) {
            int opt = val[g[x][i]]-ins[g[x][i]];
            (ret += 1LL * opt * _pow(g[x][i], K, P) % P) %= P;
            qry(g[x][i], opt); //
        }
    }
    for ( register int i = 0; i < g[x].size(); ++ i) {
        val[g[x][i]] ++;
    }
    return ret;
}

int n, x, K, P;
int main() {
    scanf("%d", &n);
    for ( register int i = 1e5; i; -- i) {
        for ( register int k = i; k <= 1e5; k += i) g[k].push_back(i);
    }
    for ( register int i = 1; i <= n; ++ i) {
        scanf("%d%d%d", &x, &K, &P);
        printf("%d\n", solve(x, K, P));
    }
    return 0;
}

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