SPOJ 5152 Brute-force Algorithm EXTREME && HDU 3221 Brute-force Algorithm 快速幂,快速求斐波那契数列,欧拉函数,同余 难度:1
5152. Brute-force Algorithm EXTREMEProblem code: BFALG |
Please click here to download a PDF version of the contest problems. The problem is problem B in the PDF. But the data limits is slightly modified: 1≤P≤1000000 in the original description, but in this EXTREME version, 1≤P≤1000000000.
=========(EDIT, Francky)===============
Professor Brute is not good at algorithm design. Once he was asked to solve a path finding problem. He worked on it for several days and finally came up with the following algorithm:
Function Find(integer n,function func) If n=1 For i = 1 to a do func() Elseif n=2 For i = 1 to b do func() Else Find(n-1,Find(n-2,func)) Function Main Find(n,funny)
Any fool but Brute knows that the function “funny” will be called too many times. Brute wants to investigate the number of times the function will be called, but he is too lazy to do it.
Now your task is to calculate how many times the function “funny” will be called, for the given a, b and n. Because the answer may be too large, you should output the answer module by P.
Input
There are multiple test cases. The first line of the input contains an integer T, meaning the number of the test cases.
For each test cases, there are four integers a, b, P and n in a single line. You can assume that 1≤n≤1000000000, 1≤P≤1000000, 0≤a, b<1000000.
Output
For each test case, output the answer with case number in a single line.
Example
Input: 3 3 4 10 3 4 5 13 5 3 2 19 100 Output: Case #1: 2 Case #2: 11 Case #3: 12
公式稍微列一下就可以发现是
次数 a,b
1: 1,0
2: 0,1
3: 1,1
4: 1,2
5: 2,3....
可以看出结果与斐波那契数列有关,
是a^f(n-3)*b^f(n-2),
但是斐波那契数列是用指数形式增长的，很快就会超出64位，而且直接运算肯定会超时，
那么
1.为了解决时间问题，使用矩阵快速幂，
{f(n-1),fn, {0,1, {fn,fn+f(n-1),
0, 0, }* 1,1, }= 0, 0}
2.为了解决斐波那契数字过大问题，有公式
a^c%P=a^(c%phi(P)+phi(P))%P
其中phi是欧拉函数
耽误时间主要原因
1 一开始想要把1-1e6所有欧拉函数值都求出来
2
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long ll;
int a,b,P,n;
ll s[2][2],t[2][2];
ll phi;
void calphisub(){
int tP=P;
phi =P;
if((tP&1)==0){
phi>>=1;
while((tP&1)==0){
tP>>=1;
}
}
for(int i=3;i*i<=tP;i+=2)
{
if(tP%i==0)
{
phi=phi/i*(i-1);
while(tP%i==0)
{
tP/=i;
}
}
}
if(tP>1)phi=phi/tP*(tP-1);
}
void multi(ll a[2][2],ll b[2][2] ,ll c[2][2] ){
ll tmp[2][2];
for(int i=0;i<2;i++){
for(int j=0;j<2;j++){
tmp[i][j]=0;
for(int k=0;k<2;k++){
tmp[i][j]+=a[i][k]*b[k][j];
if(tmp[i][j]>phi){
tmp[i][j]=tmp[i][j]%phi+phi;
}
}
}
}
for(int i=0;i<2;i++){
for(int j=0;j<2;j++){
c[i][j]=tmp[i][j];
}
}
}
void init(){
s[0][0]=1,s[0][1]=1,s[1][0]=0,s[1][1]=0;
t[0][0]=0,t[0][1]=1,t[1][0]=1,t[1][1]=1;
}
void qpow(int n){
while(n>0){
if(n%2==1){
multi(s,t,s);
}
multi(t,t,t);
n/=2;
}
}
ll qpow2(int n,ll sub){
ll ans=1;
while(n>0){
if((n&1)!=0){
ans=ans*sub%P;
}
sub=sub*sub%P;
n/=2;
}
return ans;
}
void getfab(int n,ll& fn,ll& fminus){
if(n==1){
fn=0;fminus=1;
}
else if(n==2){
fn=1;fminus=0;
}
else {
init();
qpow(n-3);
fminus=s[0][0];
fn=s[0][1];
}
}
int main(){
int T;
scanf("%d",&T);
for(int i=0;i<T;i++){
scanf("%d%d%d%d",&a,&b,&P,&n);
if(P==1){ printf("Case #%d: 0\n",i+1);continue;}
ll ta,tb;
calphisub();
getfab(n,tb,ta);
ll pa=qpow2(ta,a)%P;
ll pb=qpow2(tb,b)%P;
ll ans=pa*pb%P;
printf("Case #%d: %I64d\n",i+1,ans);
}
}