題目

分析

數論神仙題，話不多說開始推導 $ans=\sum_x^n[gcd(x,n)==1]x^d$ $=\sum_x^n\sum_{c|n,c|x}\mu(c)x^d$ $=\sum_{c|n}^n\mu(c)\sum_{x}^{\frac{n}{c}}(cx)^d$ $=\sum_{c|n}^n\mu(c)c^d\sum_{x}^{\frac{n}{c}}x^d$

結論：等冪和

$\sum_i^n i^d=\sum_i^d a_in^i$

也就是$d$次等冪和是關於$n$的$d+1$次多項式，可以採用高斯消元搞出這個玩意兒。（推導？插值啊！！） $ans=\sum_{c|n}^n\mu(c)c^d\sum_i^{d+1}a_i(\frac{n}{c})^i$ $=\sum_i^{d+1}a_i\sum_{c|n}^n\mu(c)c^d(\frac{n}{c})^i$

程式碼

#include<cstdio>
#include<algorithm>
const int P = 1e9 + 7, N = 1e3 + 10;
int ri() {
    char c = getchar(); int x = 0; for(;c < '0' || c > '9'; c = getchar()) ;
    for(;c >= '0' && c <= '9';  c = getchar()) x = (x << 1) + (x << 3) - '0' + c; return x;
}
int a[N][N], p[N], q[N], d, w;
int Pow(int x, int k) {
    k < 0 ? k += P - 1 : 0; int r = 1;
    for(;k; x = 1LL * x * x % P, k >>= 1) if(k & 1) r = 1LL * r * x % P;
    return r;
}
void Gas() {
    for(int i = 0;i <= d + 1; ++i) {
        a[i][0] = 1;
        for(int j = 1;j <= d + 1; ++j) a[i][j] = 1LL * a[i][j - 1] * (i + 1) % P;
        if(i) a[i][d + 2] = a[i - 1][d + 2];
        a[i][d + 2] = (a[i][d + 2] + a[i][d]) % P;
    }
    for(int i = 0, j; i <= d + 1; ++i) {
        for(j = i; j <= d + 1; ++j) if(a[j][i]) break;
        if(i != j) std::swap(a[i], a[j]);
        for(j = i + 1; j <= d + 1; ++j) {
            int t = 1LL * a[j][i] * Pow(a[i][i], P - 2) % P;
            for(int k = i; k <= d + 2; ++k)
                a[j][k] = (a[j][k] + P - 1LL * a[i][k] * t % P) % P;
        }
    }
    for(int i = d + 1; ~i; --i) {
        for(int j = i + 1; j <= d + 1; ++j)
            a[i][d + 2] = (a[i][d + 2] + P - 1LL * a[i][j] * a[j][d + 2] % P) % P;
        a[i][d + 2] = 1LL * a[i][d + 2] * Pow(a[i][i], P - 2) % P;
    }
}
int main() {
    d = ri(); w = ri(); Gas(); int ans = 0;
    for(int i = 1;i <= w; ++i) p[i] = ri(), q[i] = ri();
    for(int i = 0, r = 1; i <= d + 1; ans = (ans + 1LL * a[i++][d + 2] * r) % P, r = 1)
        for(int j = 1;j <= w; ++j)
            r = 1LL * r * Pow(p[j], 1LL * q[j] * i % (P - 1)) % P * (1 + P - Pow(p[j], d - i)) % P;
    printf("%d\n", ans);
    return 0;
}