BZOJ3529: [Sdoi2014]数表 莫比乌斯反演_树状数组

Code:

#include <cstdio>
#include <algorithm>
#include <cstring>
 
#define ll long long
#define setIO(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)
const long long mod = 2147483648;
const long long N = 100008;
const int maxn = 100009 ; 
 
using namespace std;
 
struct M{
    int delta;
    int id;
}opt[maxn];  
 
int prime[maxn],tot,vis[maxn],mu[maxn],g[maxn],answer[maxn];
int cmp2(M a,M b){ return a.delta < b.delta; }
void Init(){
    mu[1] = 1;
    for(int i=2;i < maxn; ++i) {
        if(!vis[i]) prime[++tot] = i, mu[i] = -1;
        for(int j=1;j <= tot && (ll)i * prime[j] <= N; ++j) {
            vis[i * prime[j]] = 1;
            if(i % prime[j]==0) {
                mu[i * prime[j]] = 0;
                break;
            }
            mu[i*prime[j]]=-mu[i];
        }
    }
    for(int i=1;i<maxn;++i)
        for(ll j=1;(ll)j*i<=N;++j) g[i*j] += i;
    for(int i=1;i<maxn;++i) opt[i].delta=g[i],opt[i].id=i;
    sort(opt+1,opt+maxn,cmp2);
}
struct BIT{
    ll C[maxn];
    int lowbit(int t) { return t & (-t); }
    void update(int x,ll k) {
        while(x < maxn) {
            C[x]+=k, C[x]%=mod;
            x+=lowbit(x);
        }
    }
    ll query(int x){
        ll sum=0;
        while(x>0) sum+=C[x],sum%=mod,x-=lowbit(x);
        return sum;
    }
}tree;
ll work(int n,int m,int p) {
    if(n>m) swap(n,m);
    long long sum=0;
    for(int i=1,j;i <= n;i=j+1) {
        j=min(n/(n/i),m/(m/i));
        sum+=(n/j)*(m/j)*(tree.query(j)-tree.query(i-1));
        sum%=mod;
    }
    return sum;
}
struct P{ int n,m,a,id; }node[maxn];
int cmp(P a,P b){ return a.a<b.a;  }
void oper(int p){ for(int i=1;i*opt[p].id<=N;++i) tree.update(opt[p].id*i,((ll)opt[p].delta*mu[i]+mod)%mod);  }
int main(){
    //setIO("input");
    Init();
    int T;
    scanf("%d",&T);
    for(int i=1;i<=T;++i) scanf("%d%d%d",&node[i].n,&node[i].m,&node[i].a),node[i].id=i;
    sort(node+1,node+1+T,cmp);
    int last=1;
    for(int i=1;i<=T;++i) {
        int j=last;
        while(opt[j].delta <= node[i].a) oper(j),++j;
        last=j;
        answer[node[i].id]=(int)work(node[i].n,node[i].m,node[i].a);
    }
    for(int i=1;i<=T;++i) printf("%d\n",(answer[i]+mod)%mod);
    return 0;
}