[BZOJ3529][Sdoi2014]数表 && 莫比乌斯反演+树状数组

来源：互联网发布：枪神纪真免费刷枪软件编辑：程序博客网时间：2024/05/22 21:55

因为这个好好地复习了一下莫比乌斯反演结果才发现这是自己的第一篇莫比乌斯反演

具体内容太长了写了两页草稿纸想看就去orz一下PoPoQQQ大爷吧

传送门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);}

0 0