Description：
To think of a beautiful problem description is so hard for me that let’s just drop them off. 😃
Given four integers
a
,
m
,
n
,
k
a,m,n,k
a
,
m
,
n
,
k
,and
S
=
g
c
d
(
a
m
−
1
,
a
n
−
1
)
%
k
S = gcd(a^{m-1},a^{n-1})\%k
S
=
g
c
d
(
a
m
−
1
,
a
n
−
1
)
%
k
,calculate the S.
Input
The first line contain a
t
t
t
,then
t
t
t
cases followed.
Each case contain four integers
a
,
m
,
n
,
k
(
1
<
=
a
,
m
,
n
,
k
<
=
10000
)
a,m,n,k(1<=a,m,n,k<=10000)
a
,
m
,
n
,
k
(
1
<
=
a
,
m
,
n
,
k
<
=
1
0
0
0
0
)
.
Output
One line with a integer S.
Sample Input
1
1 1 1 1
Sample Output
0
就是求
g
c
d
(
a
m
−
1
,
a
n
−
1
)
%
k
gcd(a^{m-1},a^{n-1})\%k
g
c
d
(
a
m
−
1
,
a
n
−
1
)
%
k
的值，这里使用了
G
C
D
GCD
G
C
D
的性质，可以先
G
C
D
GCD
G
C
D
再求幂
AC代码：
#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n, m) printf("%d %d", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sc(n) scanf("%c", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define ss(str) scanf("%s", str)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read()
{
int ret = 0, sgn = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
sgn = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
{
ret = ret * 10 + ch - '0';
ch = getchar();
}
return ret * sgn;
}
inline void Out(int a) //Êä³öÍâ¹Ò
{
if (a > 9)
Out(a / 10);
putchar(a % 10 + '0');
}
ll gcd(ll a, ll b)
{
return b == 0 ? a : gcd(b, a % b);
}
ll lcm(ll a, ll b)
{
return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(int m, int k, int mod)
{
ll res = 1, t = m;
while (k)
{
if (k & 1)
res = res * t % mod;
t = t * t % mod;
k >>= 1;
}
return res;
}
// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
return qpow(a, p - 2, p);
}
///扩展欧几里得
ll exgcd(ll a, ll b, ll &x, ll &y)
{
if (b == 0)
{
x = 1;
y = 0;
return a;
}
ll ans = exgcd(b, a % b, x, y);
ll temp = x;
x = y;
y = temp - a / b * y;
return ans;
}
///使用ecgcd求a的逆元x
ll mod_reverse(ll a, ll p)
{
ll d, x, y;
d = exgcd(a, p, x, y);
if (d == 1)
return (x % p + p) % p;
else
return -1;
}
///中国剩余定理模板
ll china(ll a[], ll b[], ll n)
{
ll a1 = a[1], b1 = b[1];
bool flag = 1;
for (int i = 2; i <= n; i++)
{
ll A = a1, B = a[i], C = b[i] - b1;
ll x, y;
ll gcd = exgcd(A, B, x, y);
if (C % gcd)
{
flag = 0;
break;
}
x = ((x * C / gcd) % (B / gcd) + (B / gcd)) % (B / gcd);
b1 = a1 * x + b1;
a1 = a1 / gcd * a[i];
}
if (b1)
return b1;
else
return a1;
}
int t;
ll a, n, m, k;
int main()
{
sd(t);
while (t--)
{
sldd(a, m);
sldd(n, k);
ll res = gcd(n, m);
ll ans = (qpow(a, gcd(m, n), k) + k - 1) % k;
pld(ans);
}
return 0;
}