【树状数组 + 容斥原理】 HDOJ 4947 GCD Array

构建辅助数组b，b[i]代表b[i]的倍数加上的数。。用树状数组维护辅助数组b的前缀和。。对于1操作，用容斥原理对非互质的数不断的加减操作。对于2操作，不断维护前缀和。。

#include <iostream>  
#include <queue>  
#include <stack>  
#include <map>  
#include <set>  
#include <bitset>  
#include <cstdio>  
#include <algorithm>  
#include <cstring>  
#include <climits>  
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 200005
#define maxm 100005
#define eps 1e-10
#define mod 998244353
#define INF 999999999
#define lowbit(x) (x&(-x))
#define mp mark_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o->ch[0], L, mid  
#define rson o->ch[1], mid+1, R  
typedef long long LL;
//typedef int LL;
using namespace std;
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}
// head

LL tree[maxn];
int pp[maxn], prime[maxn], p[maxn];
int n, m, pn;

void handle(void)
{
	pn = 0;
	for(int i = 2; i <= 200000; i++)
		if(!prime[i]) {
			for(int j = i+i; j <= 200000; j+=i)
				prime[i] = 1;
			p[pn++] = i;
		}
}
void init(void)
{
	memset(tree, 0, sizeof tree);
}
void insert(int x, int v)
{
	for(int i = x; i < maxn; i+=lowbit(i)) tree[i] += v;
}
LL query(int x)
{
	LL ans = 0;
	for(int i = x; i > 0; i-=lowbit(i)) ans += tree[i];
	return ans;
}
void dfs(int pos, int dep, int b, int add, int flag)
{
	if(b > n) return;
	if(pos == dep) {
		insert(b, add * flag);
		return;
	}
	dfs(pos+1, dep, b, add, flag);
	dfs(pos+1, dep, b * pp[pos], add, -flag);
}
void work(void)
{
	int tmp, dep, a, b, c, x, t1, t2, op;
	while(m--) {
		//scanf("%d", &op);
		scanf(op);
		if(op == 1) {
			//scanf("%d%d%d", &a, &b, &c);
			scanf(a), scanf(b), scanf(c);
			if(a % b != 0) continue;
			tmp = a / b;
			dep = 0;
			for(int i = 0; i < pn && 1LL * p[i] * p[i] <= tmp; i++)
				if(tmp % p[i] == 0) {
					while(tmp % p[i] == 0) tmp /= p[i];
					pp[dep++] = p[i];
				}
			if(tmp > 1) pp[dep++] = tmp;
			dfs(0, dep, b, c, 1);
		}
		else {
			//scanf("%d", &x);
			scanf(x);
			LL ans = 0, now, pre = 0;
			for(int i = 1; i <= x; i++) {
				t1 = x / i;
				t2 = x / t1;
				now = query(t2);
				ans += (now - pre) * t1;
				pre = now;
				i = t2;
			}
			printf("%I64d\n", ans);
		}
	}
}
int main(void)
{
	handle();
	int _ = 0;
	while(scanf("%d%d", &n, &m), n!=0 || m!=0) {
		init();
		printf("Case #%d:\n", ++_);
		work();
	}
	return 0;
}