BZOJ 3529([Sdoi2014]数表-莫比乌斯反演)

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有一张N×m的数表，其第i行第j列（1 < =i < =n，1 < =j < =m）的数值为能同时整除i和j的所有自然数之和。给定a，计算数表中不大于a的数之和模2^31的值模2^31的值。
1<=n．m<=10 5 ,Q<=2×10 4 组询问

记k的约数和f(k)

求∑ n i=1 ∑ m j=1 gcd(i,j)[f(gcd(i,j))≤a]

#include<bits/stdc++.h>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case %d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } typedef long long ll;typedef unsigned long long ull;int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} #define MAXN (100000+10)int p[MAXN],tot;bool b[MAXN]={0};int mul[MAXN]={0};pi F[MAXN]={};void make_prime(int n){    tot=0; mul[1]=1;    Fork(i,2,n)    {        if (!b[i]) p[++tot]=i,mul[i]=-1;        For(j,tot)        {            if (i*p[j]>n) break;            b[i*p[j]]=1;            mul[i*p[j]]=-mul[i];            if (i%p[j]==0) {                mul[i*p[j]]=0;                break;            }          }    }    For(i,n)        for(int j=i;j<=n;j+=i) F[j].fi+=i;    For(i,n) F[i].se=i;    sort(F+1,F+1+n);}struct comm{    int n,m,a,id;    friend bool operator<(comm a,comm b){return a.a<b.a;}}ask[MAXN];int Q;int ans[MAXN];const int N = (int)1e5+1;int t[MAXN]={0};void add(int x,int y){    for(int i=x;i<=N;i+=i&(-i)) t[i]+=y;} int query(int x) {    int tmp=0;    for(int i=x;i;i-=i&(-i)) tmp+=t[i];    return tmp;}int solve(int x) {    int tmp=0;    int n=ask[x].n,m=ask[x].m;     for(int i=1;i<=n;) {        int j=min(n/(n/i),m/(m/i));        tmp+=(n/i)*(m/i)*(query(j)-query(i-1));        i=j+1;    }    return tmp;}int main(){//  freopen("bzoj3529.in","r",stdin);//  freopen(".out","w",stdout);    make_prime(N);    Q=read();    For(i,Q) {        ask[i]=(comm){read(),read(),read(),i};         if (ask[i].n>ask[i].m) swap(ask[i].n,ask[i].m);    }     sort(ask+1,ask+1+Q);    int j=0;    For(i,Q) {        while(j<N && F[j+1].fi<=ask[i].a) {            ++j;            for(int d=F[j].se;d<=N;d+=F[j].se) add(d,F[j].fi*mul[d/F[j].se]);         }        ans[ask[i].id]=solve(i);    }    For(i,Q) printf("%d\n",ans[i]&0x7fffffff);    return 0;}

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