即求区间的乘积的欧拉函数模一个数
预处理前60个素数和逆元，用线段树维护区间乘机和每个素数出现的次数(我用了bitset嗯嗯嗯)
于是可以O(sqrt(n))求出欧拉函数
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123 123
124 /**************************************************************
125 Problem: 3813
126 User: rausen
127 Language: C++
128 Result: Accepted
129 Time:1644 ms
130 Memory:8624 kb
131 ****************************************************************/
132
133 #include <cstdio>
134 #include <bitset>
135
136 using namespace std;
137 typedef long long ll;
138 const int mod = 19961993;
139 const int N = 1e5 + 5;
140
141 struct segment{
142 ll v;
143 bitset <65> ap;
144 } seg[N << 2];
145
146 int n, m;
147 ll v;
148 bitset <65> ap;
149 int pr[65], tot;
150 ll inv[65];
151 bool vis[305];
152
153 inline int read() {
154 int x = 0;
155 char ch = getchar();
156 while (ch < ‘0‘ || ‘9‘ < ch)
157 ch = getchar();
158 while (‘0‘ <= ch && ch <= ‘9‘) {
159 x = x * 10 + ch - ‘0‘;
160 ch = getchar();
161 }
162 return x;
163 }
164
165 inline ll pow(ll x, int y) {
166 ll res = 1;
167 while (y) {
168 if (y & 1) res = res * x % mod;
169 x = x * x % mod;
170 y >>= 1;
171 }
172 return res;
173 }
174
175 void pre_work() {
176 int i, j;
177 for (i = 2; tot <= 60; ++i)
178 if (!vis[i]) {
179 pr[++tot] = i;
180 for (j = i; j <= 300; j += i) vis[j] = 1;
181 }
182 for (i = 1; i <= 60; ++i)
183 inv[i] = 1ll * pow(pr[i], mod - 2) * (pr[i] - 1) % mod;
184 }
185
186 #define Ls (p << 1)
187 #define Rs (p << 1 | 1)
188 #define mid (l + r >> 1)
189 inline void update(int p) {
190 seg[p].v = seg[Ls].v * seg[Rs].v % mod;
191 seg[p].ap = seg[Ls].ap | seg[Rs].ap;
192 }
193
194 void seg_build(int p, int l, int r) {
195 if (l == r) {
196 seg[p].v = 3, seg[p].ap[2] = 1;
197 return;
198 }
199 seg_build(Ls, l, mid), seg_build(Rs, mid + 1, r);
200 update(p);
201 }
202
203 void seg_modify(int p, int l, int r, int pos, int v) {
204 if (l == r) {
205 int i;
206 seg[p].v = v, seg[p].ap.reset();
207 for (i = 1; i <= 60; ++i)
208 if (v % pr[i] == 0) seg[p].ap[i] = 1;
209 return;
210 }
211 if (pos <= mid) seg_modify(Ls, l, mid, pos, v);
212 else seg_modify(Rs, mid + 1, r, pos, v);
213 update(p);
214 }
215
216 void seg_query(int p, int l, int r, int L, int R) {
217 if (L <= l && r <= R) {
218 v = v * seg[p].v % mod, ap |= seg[p].ap;
219 return;
220 }
221 if (L <= mid) seg_query(Ls, l, mid, L, R);
222 if (mid < R) seg_query(Rs, mid + 1, r, L, R);
223 }
224 #undef mid
225 #undef Ls
226 #undef Rs
227
228 int main() {
229 int oper, x, y, i;
230 pre_work();
231 n = 1e5;
232 seg_build(1, 1, n);
233 m = read();
234 while (m--) {
235 oper = read(), x = read(), y = read();
236 if (oper) seg_modify(1, 1, n, x, y);
237 else {
238 v = 1, ap.reset();
239 seg_query(1, 1, n, x, y);
240 for (i = 1; i <= 60; ++i)
241 if (ap[i]) v = v * inv[i] % mod;
242 printf("%lld\n", v);
243 }
244 }
245 return 0;
246 }
时间： 2024-12-23 05:32:31