【树状数组 + 容斥原理】 HDOJ 4947 GCD Array

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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();}// headLL 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;}

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