[BZOJ3529][Sdoi2014]數表 && 莫比烏斯反演+樹狀數組

因爲這個好好地複習了一下莫比烏斯反演 結果才發現這是自己的第一篇莫比烏斯反演

具體內容太長了 寫了兩頁草稿紙 想看就去orz一下PoPoQQQ大爺吧

傳送門http://blog.csdn.net/popoqqq/article/details/42076231

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
#define SF scanf
#define PF printf
#define mp make_pair
#define fir first
#define sec second
#define lowbit(x) ((x) & (-(x)))
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int MAXN = 100000;
const int MOD = 2147483647;
int mu[MAXN+10], prime[MAXN+10], tot, ans[MAXN+10], c[MAXN+10], MAX;
pii F[MAXN+10];
bool vis[MAXN+10];
int Q;
struct Node {
    int n, m, a, id;
    bool operator < (const Node &t) const {
        return a < t.a;
    }
} A[MAXN+10];
void init() {
    mu[1] = 1;
    for(int i = 2; i <= MAX; i++) {
        if(!vis[i]) prime[++tot] = i, mu[i] = -1;
        for(int j = 1; prime[j]*i <= MAX; j++) {
            vis[prime[j]*i] = 1;
            if(i % prime[j] == 0) {
                mu[prime[j]*i] = 0;
                break;
            }
            mu[prime[j]*i] = -mu[i];
        }
    }
    for(int i = 1; i <= MAX; i++) {
        F[i].sec = i;
        for(int j = i; j <= MAX; j += i)
            F[j].fir += i;
    }
}
void add(int x, int v) {
    for( ; x <= MAX; x += lowbit(x)) c[x] += v;
}
int query(int x) {
    int ret = 0;
    for( ; x; x -= lowbit(x)) ret += c[x];
    return ret;
}
void calc(int i) {
    int x = A[i].id, n = A[i].n, m = A[i].m, Next_pos;
    for(int i = 1; i <= n; i = Next_pos+1) {
        Next_pos = min(n / (n/i), m / (m/i));
        ans[x] += (n/i) * (m/i) * (query(Next_pos) - query(i-1));
    }
}
int main() 
{
     
    SF("%d", &Q);
    for(int i = 1; i <= Q; i++) {
        SF("%d%d%d", &A[i].n, &A[i].m, &A[i].a);
        if(A[i].n > A[i].m) swap(A[i].n, A[i].m);
        A[i].id = i; MAX = max(A[i].m, MAX);
    }
    init();
    sort(A+1, A+1+Q);
    sort(F+1, F+1+MAX);
    int cur = 0;
    for(int i = 1; i <= Q; i++) {
        while(cur < MAX && F[cur+1].fir <= A[i].a) {
            cur++;
            for(int j = F[cur].sec; j <= MAX; j += F[cur].sec) 
                add(j, F[cur].fir * mu[j / F[cur].sec]);
        }
        calc(i);
    }
    for(int i = 1; i <= Q; i++) PF("%d\n", ans[i]&MOD);
}