hdu 6363 bookshelf

题解讲的很清楚了，直接看代码就懂了

题解:http://bestcoder.hdu.edu.cn/blog/2018-multi-university-training-contest-6-solutions-by-%e7%a6%8f%e5%b7%9e%e5%a4%a7%e5%ad%a6/

#include<stdio.h>
#include<iostream>
#include<vector>
#include<string.h>
#include<algorithm>
#include<map>
#include<queue>
#include<cmath>
#include<set>
#include<map>
#define de(x) cout<<#x<<"="<<x<<endl;
#define dd(x) cout<<#x<<"="<<x<<" ";
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define repd(i,a,b) for(int i=a;i>=(b);--i)
#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)
#define mt(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define mp(u,v) make_pair(u,v)
#define sz(a) (int)a.size()
#define pb push_back
#define PI acos(-1.0)
#define qc std::ios::sync_with_stdio(false)
#define all(a) a.begin(),a.end()
using namespace std;
typedef vector<int> vi;
typedef long long ll;
typedef double db;
typedef pair<int,int> pii;

const ll mod = 1e9+;
const int N = 2e6+;
const double eps = 1e-;
const int inf = 0x3f3f3f3f;

bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }
bool ls(const db &a, const db &b) { return a + eps < b; }
bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }
ll gcd(ll a,ll b) { return a==?b:gcd(b%a,a); };
ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }
ll kpow(ll a,ll b) {ll res=;a%=mod; if(b<) return ; for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}
ll read(){
    ll x=,f=;char ch=getchar();
    while (ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
ll inv[N], f[N], fac[N];
ll a[N];
void init() {
    f[] = ; f[] = ;
    fac[] = fac[] = ;
    inv[]=;
    rep(i, , N) {
        fac[i] = fac[i-] * (ll)i % mod;
        inv[i] = kpow(fac[i], mod-);
        f[i] = (f[i-] * f[i-]) % mod;
    }
}
ll C(int n, int m) {
    if(n < m) return 0ll;
    if(m== || n==m) return 1ll;
    if(n-==m || m==) return n;
    return fac[n] * inv[m] % mod * inv[n-m] % mod;
}
int main(){
    init();
    int T;
    scanf("%d",&T);
    while(T--) {
        int n, m;
        scanf("%d%d",&n,&m);
        mt(a, );
        ll sum = C(n+m-, n), ans = ;
        repd(g,n,) {
            if(n%g==) {
                ll j = n / g;
                a[g] = C(j+m-,j) % mod;
                repp(k,g+g,n + ,g) a[g] = (a[g] - a[k] + mod) % mod;
                ans = ans + a[g] * (f[g] -  + mod) % mod; ans %= mod;
            }
        }
        ans = ans * kpow(sum, mod-) % mod;
        printf("%lld\n",ans);
    }
    return ;
}