BZOJ3601 一个人的数论

Posted 2020-09-17

BZOJ3601 一个人的数论

题意

\\[ ans = \\sum _ {i \\nmid n} i ^ d \\]

\\[ n = \\prod _ {i = 1} ^ {w} p _ i ^ {\\alpha _ i} \\]

输入

第一行给出$d$, $w$

接下来$w$行，第$i + 1$行给出$p _ i$, $\\alpha _ i$

输出

题目所求的$ans$

样例输入

3 2
2 1
5 1

样例输出

1100

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 围观大神题解：https://www.cnblogs.com/jianglangcaijin/p/4033399.html

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 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 template <class _T> inline void read(_T &_x) {
 4     int _t; bool flag = false;
 5     while ((_t = getchar()) != ‘-‘ && (_t < ‘0‘ || _t > ‘9‘)) ;
 6     if (_t == ‘-‘) _t = getchar(), flag = true; _x = _t - ‘0‘;
 7     while ((_t = getchar()) >= ‘0‘ && _t <= ‘9‘) _x = _x * 10 + _t - ‘0‘;
 8     if (flag) _x = -_x;
 9 }
10 typedef long long LL;
11 const int maxw = 1010;
12 const int maxd = 110;
13 const int mod = 1e9 + 7;
14 inline int mul(int a, int b) {return (int)((LL)a * b % mod); }
15 inline int add(int a, int b) {
16     a += b;
17     if (a < 0) a += mod; if (a >= mod) a -= mod;
18     return a;
19 }
20 int qpow(LL a, LL b) {
21     if (b < 0) return qpow(qpow(a, mod - 2), -b);
22     LL ret = 1;
23     while (b) {
24         if (b & 1) (ret *= a) %= mod;
25         (a *= a) %= mod, b >>= 1;
26     }
27     return (int)ret;
28 }
29 int H(int i, int p, int a, int d) {
30     return mul(qpow(p, (LL)a * i), add(1, -qpow(p, d - i)));
31 }
32 int A[maxd][maxd], X[maxd];
33 inline void init(int d) {
34     static int sum[maxd];
35     sum[0] = 0;
36     for (int i = 1; i <= d + 2; ++i) {
37         sum[i] = add(sum[i - 1], qpow(i, d));
38         A[i - 1][d + 2] = sum[i], A[i - 1][0] = 1;
39         for (int j = 1; j <= d + 1; ++j) A[i - 1][j] = mul(A[i - 1][j - 1], i);
40     }
41     for (int i = 0, tmp; i <= d + 1; ++i) {
42         int to = i;
43         while (to <= d + 1 && !A[to][i]) break;
44         if (i != to) swap(A[i], A[to]);
45         for (int j = 0; j <= d + 1; ++j) if (j != i && A[j][i]) {
46            tmp = mul(A[j][i], qpow(A[i][i], mod - 2));
47            for (int k = 0; k <= d + 2; ++k) A[j][k] = add(A[j][k], -mul(tmp, A[i][k]));
48         }
49     }
50     for (int i = 0; i <= d + 1; ++i) {
51         X[i] = mul(A[i][d + 2], qpow(A[i][i], mod - 2));
52     }
53 }
54 int d, w, p[maxw], a[maxw];
55 int main() {
56     //freopen(".in", "r", stdin);
57     //freopen(".out", "w", stdout);
58     read(d), read(w);
59     for (int i = 1; i <= w; ++i) read(p[i]), read(a[i]);
60     init(d);
61     int ans = 0;
62     for (int i = 1; i <= d + 1; ++i) {
63         int ret = X[i];
64         for (int j = 1; j <= w; ++j)
65             ret = mul(ret, H(i, p[j], a[j], d));
66         ans = add(ans, ret);
67     }
68     cout << ans << endl;
69     return 0;
70 }

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