四一。 数论 。4.19（gcd 扩展gcd，素数 ，快速幂)
A 青蛙 POJ 1061 扩展gcd
#include
#include
#include
using namespace std;
long long extgcd(long long a,long long b,long long &x,long long &y)
{
int d = a;
if(b != 0)
{
d = extgcd(b, a%b, y, x);
y-= (long long)(a/b) * x;
}
else
{
x = 1;
y = 0;
}
//cout<<" x :"<>x>>y>>m>>n>>l;
long long c=y-x; ///ax+by = c
long long gcd,temp;
gcd = extgcd(m-n,l,p,q); ///gcd 最大公约数
if(c % gcd != 0) ///c不为gcd的倍数则无解
cout<<"Impossible"< using namespace std;
bool is_prime[10005];/// is_prime[i]为true表示i为素数
// 返回n以内素数的个数
int sieve(int n)
{
int p = 0;
for(int i = 2; i <= n; i++)
is_prime[i] = true;
is_prime[0] = is_prime[1] = false;
for(int i = 2; i <= n; i++)
{
if(is_prime[i])
{
prime[p++] = i;
for(int j = 2*i; j <= n; j += i)
is_prime[j] = false;
}
}
return p;
}
int opt(int temp)
{
int n = temp/2,num=0;
//cout<
if(is_prime[temp-prime[i]]) /// 枚举 temp减去一个为素数的因子判断余数是否为素数
{
num++;
//cout<
}
return num;
}
int main()
{
int temp;
int p =sieve(10000);
while(cin>>temp)
{
if(temp == 0)
return 0;
else
{
cout<
}
return 0;
}
C 周期 扩展gcd
#include
using namespace std;
long long extgcd(long long a,long long b,long long &x,long long &y)
{
long long d = a;
if(b != 0)
{
d = extgcd(b, a%b, y, x);
y-= (long long)(a/b) * x;
}
else
{
x = 1;
y = 0;
}
//cout<<" x :"<
}
int main()
{
long long p,e,i,d;
int n=1;
while(cin>>p>>e>>i>>d)
{
if(p == e && e == i && i == d && d == -1)
return 0;
{
long long c = e-p,l;
long long temp,sum,gcd,x,y;
gcd = extgcd(23,28,x,y); ///gcd 最大公约数
//cout<
sum = p + x*23;
sum = (sum%644 + 644)%644; /// ax+by = c, 求最小非负c，作为新一方程的常数项
gcd = extgcd(23*28/gcd,33,x,y);
c = i - sum;
x = (long long)x*c/gcd;
sum = (23*28/gcd*x+sum);
sum = (sum%21252+21252)%21252; /// 最小非负c
if(sum <= d) /// 坑： sum必须大于一个周期 否则例一为0，例二为-10
sum+=21252;
cout<<"Case "<
}
return 0;
}
D POJ 1995 幂和 快速幂
#include
using namespace std;
typedef long long ll;
ll mod_pow(ll x, ll n, ll mod)
{
if(n == 0)
return 1;
ll res = mod_pow(x * x % mod, n / 2, mod); /// 递归直到0次方
if(n & 1) /// n二进制与1二进制得数取最后一位（为奇数）【n为奇数要*x】
res = res * x % mod;
return res;
}
int main()
{
int n,mod;
cin>>n;
while(n--)
{
int m, sum = 0, x, y;
cin>>mod;
cin>>m;
while(m--)
{
cin>>x>>y;
sum = sum + mod_pow(x, y, mod); /// 多项和取幂等于分别取幂再取幂
}
cout << sum % mod << endl;
}
return 0;
}
E 最小公倍数 gcd
using namespace std;
int gcd(int a, int b)
{
if( b == 0)
return a;
else
gcd(b, a%b);
}
int lcm(int a, int b)
{
return (long long)a*b/gcd(a,b);
int main()
{
int n;
while(cin>>n)
{
int x,y;
cin>>x;
n = n-1;
while(n--)
{
cin>>y;
x=lcm(x,y); /// 两个两个求lcm
}
cout<
return 0;
}