模幂运算
public static int modPowerShow(int a, int k, int n){
System.out.println("*************************************************************************");
System.out.println("****NOTICE:[\"\\\" area should be replaced by blank so there is nothing]****");
System.out.println("****NOTICE:[\"(3=))\" area should be replaced by three line equal]*********");
System.out.println("*************************************************************************");
System.out.println("i\t\tki\t\tx\t\ty");
int now, x = 1, y = a, power;
for(int i = 0;;i++){
now = k % 2;
System.out.print(i + "\t\t" + now + "\t\t");
power = (x * y) % n;
if(now == 1) {
System.out.print(x + "x" + y + "(3=)" + power + "\t\t");
x = power;
}
else System.out.print(x + "\t\t");
y = (y * y) % n;
System.out.println(a + "^(2^" + (i+1) + ")" + "(3=)" + y);
k /= 2;
if(k == 1)break;
}
power = (x * y) % n;
System.out.println("\\\t\t\\\t\t" + x + "x" + y + "(3=)" + power);
return power;
}
模逆运算
public static int modInverseShow(int d, int n){
System.out.println("*************************************************************************");
System.out.println("****NOTICE:[\"\\\" area should be replaced by blank so there is nothing]****");
System.out.println("*************************************************************************");
int up = n;
int down = d;
int inverseUp = 0;
int inverse = 1;
int times, temp;
System.out.println("\\\t\tu\t\tv\t\tq");
System.out.println(up + "\t\t1\t\t0\t\t\\");
System.out.print(down + "\t\t0\t\t1\t\t");
for(;;) {
times = up / down;
System.out.println(times);
temp = up - down * times;
up = down;
down = temp;
temp = inverseUp - inverse * times;
inverseUp = inverse;
inverse = temp;
System.out.print(down + "\t\t\\\t\t" + inverse + "\t\t");
if(down == 1)return (inverse > 0) ? inverse : (inverse + n);
if(down <= 0)return -1;
}
}
注意：如果返回值为-1，可能出现意外情况（两数不互素等原因），请自行验算
