Brute-force Algorithm
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2560 Accepted Submission(s): 657
Problem Description
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:
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.
Sample Input
3
3 4 10 3
4 5 13 5
3 2 19 100
Sample Output
Case #1: 2
Case #2: 11
Case #3: 12
Source
2009 Asia Shanghai Regional Contest Host by DHU
/**
题意：根据题意可以知道求 f(n) = f(n-1)*f(n-2)的值
f(1) = a
f(2) = b;
f(3) = a*b;
f(4) = a*b^2
f(5) = a^2*b^3
......
可以得知 a,b 的指数是斐波纳锲数列
做法：欧拉 + 矩阵 + 蒙哥马利幂模算法 ps:没有搞清楚一点 在求矩阵的n-3次幂可以过 可是n-2次幂 不能过
**/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
#define SIZE 2
#define clr( a, b ) memset( a, b, sizeof(a) )
long long MOD;
struct Mat
{
long long mat[ SIZE ][ SIZE ];
int n;
Mat(int _n) {
n = _n;
clr(mat, 0);
}
void init() {
for(int i = 0; i < n; ++i)
for(int j = 0; j < n; ++j) {
mat[i][j] = (i == j);
}
}
Mat operator * (const Mat& b) const {
Mat c(b.n);
for(int k = 0; k < n; ++k)
for(int i = 0; i < n; ++i) {
if(mat[i][k])
for(int j = 0; j < n; ++j) {
c.mat[i][j] = (c.mat[i][j] + mat[i][k] * b.mat[k][j]) % MOD;
}
}
return c;
}
};
Mat fast_mod(Mat a, int b)
{
Mat res(a.n);
res.init();
while(b)
{
if(b & 1) {
res = res * a;
}
a = a * a;
b >>= 1;
}
return res;
}
long long eular(long long n)
{
long long ans = n;
for(int i = 2; i * i <= n; i++)
{
if(n % i == 0)
{
ans -= ans / i;
while(n % i == 0) {
n /= i;
}
}
}
if(n > 1) {
ans -= ans / n;
}
return ans;
}
long long modPow(long long s, long long index, long long mod)
{
long long ans = 1;
s %= mod;
while(index >= 1)
{
if((index & 1) == 1) { //奇数
ans = (ans * s) % mod;
}
index >>= 1;
s = s * s % mod;
}
return ans;
}
int main()
{
int T;
int Case = 1;
scanf("%d", &T);
while(T--)
{
long long x, y, n, res, p;
cin >> x >> y >> p >> n;
printf("Case #%d: ", Case++);
if(p == 0 || p == 1) {
printf("0\n");
continue;
}
if(n == 1)
{
cout << x % p << endl;
continue;
}
else if(n == 2)
{
cout << y % p << endl;
continue;
}
MOD = eular(p);
Mat C(2);
C.mat[0][0] = 0;
C.mat[0][1] = 1;
C.mat[1][0] = 1;
C.mat[1][1] = 1;
C = fast_mod(C, n - 3);
long long aa = C.mat[0][0] + C.mat[0][1];
long long bb = C.mat[1][1] + C.mat[1][0];
res = ((modPow(x , aa, p) * modPow(y , bb, p)) % p + p) % p;
cout << res << endl;
}
return 0;
}
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
#define SIZE 2
#define clr( a, b ) memset( a, b, sizeof(a) )
long long MOD;
struct Mat
{
long long mat[ SIZE ][ SIZE ];
int n;
Mat(int _n) {
n = _n;
clr(mat, 0);
}
void init() {
for(int i = 0; i < n; ++i)
for(int j = 0; j < n; ++j) {
mat[i][j] = (i == j);
}
}
Mat operator * (const Mat& b) const {
Mat c(b.n);
for(int k = 0; k < n; ++k)
for(int i = 0; i < n; ++i) {
if(mat[i][k])
for(int j = 0; j < n; ++j) {
c.mat[i][j] = (c.mat[i][j] + mat[i][k] * b.mat[k][j]) % MOD;
}
}
return c;
}
};
Mat fast_mod(Mat a, int b)
{
Mat res(a.n);
res.init();
while(b)
{
if(b & 1) {
res = res * a;
}
a = a * a;
b >>= 1;
}
return res;
}
long long eular(long long n)
{
long long ans = n;
for(int i = 2; i * i <= n; i++)
{
if(n % i == 0)
{
ans -= ans / i;
while(n % i == 0) {
n /= i;
}
}
}
if(n > 1) {
ans -= ans / n;
}
return ans;
}
long long modPow(long long s, long long index, long long mod)
{
long long ans = 1;
s %= mod;
while(index >= 1)
{
if((index & 1) == 1) { //奇数
ans = (ans * s) % mod;
}
index >>= 1;
s = s * s % mod;
}
return ans;
}
int main()
{
int T;
int Case = 1;
scanf("%d", &T);
while(T--)
{
long long x, y, n, res, p;
cin >> x >> y >> p >> n;
printf("Case #%d: ", Case++);
if(p == 0 || p == 1) {
printf("0\n");
continue;
}
if(n == 1)
{
cout << x % p << endl;
continue;
}
else if(n == 2)
{
cout << y % p << endl;
continue;
}
MOD = eular(p);
Mat C(2);
C.mat[0][0] = 1;
C.mat[0][1] = 1;
C.mat[1][0] = 1;
C.mat[1][1] = 0;
C = fast_mod(C, n - 3);
long long aa = C.mat[1][0] + C.mat[1][1];
long long bb = C.mat[0][0] + C.mat[0][1];
res = ((modPow(x , aa, p) * modPow(y , bb, p)) % p + p) % p;
cout << res << endl;
}
return 0;
}