# |
Submission time |
Handle |
Problem |
Language |
Result |
Execution time |
Memory |
15437 |
2015-07-12T07:39:58 Z |
tonyjjw |
괄호 문자열 (kriii3_R) |
C++14 |
|
408 ms |
71796 KB |
#include<stdio.h>
#include<vector>
#define ML 1000000
#pragma warning(disable:4996)
using namespace std;
typedef long long ll;
int p,q,mod,Q;
ll pow2[ML+1];
ll A[ML+1];
vector<ll> C[ML+1];
long long gcd(long long a,long long b) {
if(b == 0) return a;
return gcd(b,a % b);
}
pair<long long,long long> extended_gcd(long long a,long long b) {
if(b == 0) return make_pair(1,0);
pair<long long,long long> t = extended_gcd(b,a % b);
return make_pair(t.second,t.first - t.second * (a / b));
}
long long modinverse(long long a,long long m) {
return (extended_gcd(a,m).first % m + m) % m;
}
long long chinese_remainder_two(long long a1,long long n1,long long a2,long long n2)
{
return ( a1*n2*modinverse(n2,n1) + a2 * n1 * modinverse(n1,n2) ) % (n1*n2);
}
pair<long long,long long> chinese_remainder(vector<long long> a,vector<long long> n) {
pair<long long,long long> res(a[0],n[0]);
for(int i = 1; i < a.size(); i++){
auto g = gcd(res.second,n[i]);
if((a[i] - res.first) % g) {
res = make_pair(-1,-1);
break;
}
res.first = chinese_remainder_two(res.first / g,res.second / g,a[i] / g,n[i] / g) * g + a[i] % g;
res.second = res.second / g * n[i];
res.first %= res.second;
}
return res;
}
int main(){
scanf("%d%d%d%d",&p,&q,&mod,&Q);
if(p==1 && q==0){ //subtask 1
vector<ll> factor;
int L;
for(int i=2;i*i<=mod;i++){
if(mod%i)continue;
int v=1;
while(mod%i==0)mod/=i,v*=i;
factor.push_back(v);
}
if(mod!=1)factor.push_back(mod);
for(L=1,pow2[0]=1%mod;L<=ML;L++){
pow2[L]=pow2[L-1]*2%mod;
}
for(auto mod:factor){
C[1].push_back(1);
for(L=2;L<=ML;L++){
C[L].push_back(2*C[L-1].back()*(2*L-1)%mod*modinverse(L+1,mod)%mod);
}
}
for(int i=0;i<Q;i++){
scanf("%d",&L);
if(L&1){
puts("0");
continue;
}
ll v=pow2[L]-chinese_remainder(C[L/2],factor).first;
v%=mod;
if(v<0)v+=mod;
printf("%lld\n",v);
}
}
else if(p==0 && q==1){ //subtask 2
int L;
for(L=1,pow2[0]=1%mod;L<=ML;L++){
pow2[L]=pow2[L-1]*2%mod;
}
for(A[1]=0,A[2]=2%mod,L=1;L<=ML-2;L++){
int m=L/2+1;
A[L+2]=4*A[L]+(pow2[m]-A[m])*pow2[L+2-2*m];
A[L+2]%=mod;
if(A[L+2]<0)A[L+2]+=mod;
}
for(int i=0;i<Q;i++){
scanf("%d",&L);
ll v=pow2[L]-A[L];
v%=mod;
if(v<0)v+=mod;
printf("%lld\n",v);
}
}
return 0;
}
# |
Verdict |
Execution time |
Memory |
Grader output |
1 |
Incorrect |
408 ms |
71796 KB |
Output isn't correct |
2 |
Halted |
0 ms |
0 KB |
- |
# |
Verdict |
Execution time |
Memory |
Grader output |
1 |
Correct |
66 ms |
40644 KB |
Output is correct |
2 |
Correct |
76 ms |
40644 KB |
Output is correct |
3 |
Correct |
82 ms |
40644 KB |
Output is correct |
4 |
Correct |
80 ms |
40644 KB |
Output is correct |
5 |
Correct |
87 ms |
40644 KB |
Output is correct |
6 |
Correct |
75 ms |
40644 KB |
Output is correct |
# |
Verdict |
Execution time |
Memory |
Grader output |
1 |
Incorrect |
8 ms |
40640 KB |
Output isn't correct |
2 |
Halted |
0 ms |
0 KB |
- |