T2 扩展BSGS
T3 快速阶乘
给定整数n，质数p和正整数c，求整数s和b，满足n! / pb = s mod pc
考虑每次取出floor(n/p)个p因子，然后将问题转化为子问题。
1 /**************************************************************
2 Problem: 3283
3 User: idy002
4 Language: C++
5 Result: Accepted
6 Time:1704 ms
7 Memory:12380 kb
8 ****************************************************************/
9
10 #include <cstdio>
11 #include <cstring>
12 #include <cmath>
13
14 typedef long long dnt;
15 void exgcd( dnt a, dnt b, dnt &d, dnt &x, dnt &y ) {
16 if( b==0 ) {
17 x=1, y=0, d=a;
18 } else {
19 exgcd(b,a%b,d,y,x);
20 y-=a/b*x;
21 }
22 }
23 dnt inv( int a, int n ) {
24 dnt d, x, y;
25 exgcd(a,n,d,x,y);
26 return d==1 ? (x%n+n)%n : -1;
27 }
28 dnt mpow( dnt a, dnt b, dnt c ) {
29 dnt rt;
30 for( rt=1; b; b>>=1,a=a*a%c )
31 if( b&1 ) rt=rt*a%c;
32 return rt;
33 }
34
35 namespace Task1 {
36 void sov( dnt a, dnt b, dnt c ) {
37 printf( "%lld\n", mpow(a,b,c) );
38 }
39 };
40 namespace Task2 {
41 const int mod = 38281;
42 const int len = mod<<3;
43 struct Hash {
44 int head[mod], val[len], rat[len], next[len], ntot;
45 void init() {
46 ntot=0;
47 memset( head, 0, sizeof(head) );
48 }
49 void insert( int v, int r ) {
50 int k = v % mod;
51 ntot++;
52 next[ntot] = head[k];
53 val[ntot] = v;
54 rat[ntot] = r;
55 head[k] = ntot;
56 }
57 int query( int v ) {
58 int k = v % mod;
59 for( int t=head[k]; t; t=next[t] )
60 if( val[t]==v ) return rat[t];
61 return -1;
62 }
63 }hash;
64
65 dnt gcd( dnt a, dnt b ) {
66 return b ? gcd(b,a%b) : a;
67 }
68 int bsgs( dnt s, dnt a, dnt b, dnt c ) {
69 hash.init();
70 int m = ceil(sqrt(c));
71 for( int i=0; i<m; i++ ) {
72 if( s==b ) return i;
73 hash.insert( s, i );
74 s = s*a % c;
75 }
76 dnt am = 1;
77 for( int i=0; i<m; i++ )
78 am = am*a % c;
79 am = inv(am,c);
80 b = b*am % c;
81 for( int i=m; i<c; i+=m ) {
82 int j = hash.query( b );
83 if( j!=-1 ) return i+j;
84 b = b*am % c;
85 }
86 return -1;
87 }
88 int exbsgs( dnt a, dnt b, dnt c ) {
89 dnt s = 1;
90 for( int i=0; i<32; i++ ) {
91 if( s==b ) return i;
92 s = s*a % c;
93 }
94 dnt cd;
95 s = 1;
96 int rt = 0;
97 while( (cd=gcd(a,c))!=1 ) {
98 rt++;
99 s*=a/cd;
100 if( b%cd ) return -1;
101 b/=cd;
102 c/=cd;
103 s%=c;
104 }
105 int p = bsgs(s,a,b,c);
106 if( p==-1 ) return -1;
107 return rt + p;
108 }
109 void sov( int a, int b, int c ) {
110 int res = exbsgs(a,b,c);
111 if( res==-1 ) printf( "Math Error\n" );
112 else printf( "%d\n", res );
113 }
114 };
115 namespace Task3 {
116 struct Pair {
117 int s, k;
118 Pair( int s, int k ):s(s),k(k){}
119 };
120 dnt aa[50], mm[50];
121 dnt pres[1001000];
122 dnt pp[50], cc[50], ppp[50], tot;
123
124 dnt china( int n, dnt *a, dnt *m ) {
125 int M=1;
126 for( int i=0; i<n; i++ )
127 M *= m[i];
128 int rt = 0;
129 for( int i=0; i<n; i++ ) {
130 dnt Mi = M/m[i];
131 rt = (rt+Mi*inv(Mi,m[i])*a[i]) % M;
132 }
133 return rt;
134 }
135 void init( int p, int pp ) {
136 pres[0] = 1;
137 for( int i=1; i<=pp; i++ ) {
138 if( i%p==0 ) {
139 pres[i] = pres[i-1];
140 } else {
141 pres[i] = pres[i-1]*i % pp;
142 }
143 }
144 }
145 Pair split( int n, int p, int c, int pp ) {
146 int b = n/p;
147 if( b==0 ) {
148 return Pair( pres[n], 0 );
149 } else {
150 Pair pr = split( b, p, c, pp );
151 return Pair( (pr.s*pres[n%pp]%pp) * mpow(pres[pp],n/pp,pp) % pp, pr.k+b );
152 }
153 }
154 void sov( int m, int n, int c ) {
155 tot = 0;
156 for( int i=2; i*i<=c; i++ ) {
157 if( c%i==0 ) {
158 pp[tot] = i;
159 cc[tot] = 0;
160 ppp[tot] = 1;
161 while( c%i==0 ) {
162 cc[tot]++;
163 ppp[tot] *= pp[tot];
164 c/=i;
165 }
166 tot++;
167 }
168 }
169 if( c!=1 ) {
170 pp[tot] = c;
171 cc[tot] = 1;
172 ppp[tot] = c;
173 tot++;
174 c = 1;
175 }
176 for( int i=0; i<tot; i++ ) {
177 init(pp[i],ppp[i]);
178 Pair pn = split( n, pp[i], cc[i], ppp[i] );
179 Pair pa = split( m, pp[i], cc[i], ppp[i] );
180 Pair pb = split( n-m, pp[i], cc[i], ppp[i] );
181 if( pn.k-pa.k-pb.k >= cc[i] ) {
182 aa[i] = 0;
183 mm[i] = ppp[i];
184 } else {
185 aa[i] = pn.s * (inv(pa.s,ppp[i])*inv(pb.s,ppp[i])%ppp[i]) % ppp[i];
186 for( int j=0; j<pn.k-pa.k-pb.k; j++ )
187 aa[i] = (dnt) aa[i]*pp[i] % ppp[i];
188 mm[i] = ppp[i];
189 }
190 }
191 /*
192 fprintf( stderr, "tot=%d\n", tot );
193 for( int i=0; i<tot; i++ )
194 fprintf( stderr, "%d %d\n", aa[i], mm[i] );
195 */
196 printf( "%lld\n", china(tot,aa,mm) );
197 }
198 };
199
200 int main() {
201 int n;
202 scanf( "%d", &n );
203 for( int i=1,opt,y,z,p; i<=n; i++ ) {
204 scanf( "%d%d%d%d", &opt, &y, &z, &p );
205 if( opt==1 )
206 Task1::sov( y, z, p );
207 else if( opt==2 )
208 Task2::sov( y, z, p );
209 else
210 Task3::sov( y, z, p );
211 }
212 }
时间： 2024-10-19 14:08:17