【树状数组 + 容斥原理】 HDOJ 4947 GCD Array
2014-08-16 10:07
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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();} // 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; }
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