【Java】(有步骤!)模逆、模幂、十进制转十六进制、十六进制转十进制、xTime算法、LFSR画状态图、椭圆曲线加法、椭圆曲线乘法、获得椭圆曲线上的点

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功能表:

public class LFSR {
    private int state;
    private String structConst;
    private int n;

    public LFSR(int state, String structConst) {
        this.state = state;
        this.structConst = structConst;
        this.n = structConst.length();

        drawState();
    }

    public int getState() {
        return state;
    }


    private int go() {
        int out = state;
        out >>= n-1;

        int in = 0;
        for(int i = 0; i < n; i++){
            if(structConst.charAt(i) == '1') {
                in = in^((state>>i)%2);
            }
        }

        state <<= 1;
        state %= 1 << n;
        state += in;

        return out;
    }

    public void drawState(){
        System.out.print("状态图:");
        int[] states = new int[1<<n];
        LFSR srTemp = this;
        boolean flag = true;
        int num = 0;
        for(int i = 0; flag; i++){
            states[i] = srTemp.getState();
            srTemp.go();
            for(int j = 0; j < i; j++){
                if (srTemp.getState() == states[j]) {
                    flag = false;
                }
            }
            num = i;
        }
        for(int i = 0; i < num + 1; i++){
            for(int j = 0; j < n; j++){
                int out = (states[i]>>n-j-1) % 2;
                System.out.print(out);
            }
            System.out.print("→");
        }
        System.out.println("接开头循环");
    }
}
public class Exam20200622 {

    public static int modInverse(int d, int n){
        int a, b, q, r, u=0, v=1, t;
        a = n;
        b = (d>=0) ? (d%n) : -(d%n);

        while(b != 0){
            q = a / b;
            r = a - b * q;
            a = b;
            b = r;
            t = v;
            v = u - q * v;
            u = t;
        }

        if(a != 1)return -a;
        else return (u<0) ? u+n : u;
    }

    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("\		u		v		q");
        System.out.println(up + "		1		0		\");
        System.out.print(down + "		0		1		");
        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 + "		\		" + inverse + "		");
            if(down == 1)return (inverse > 0) ? inverse : (inverse + n);
            if(down <= 0)return -1;
        }
    }

    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		ki		x		y");
        int now, x = 1, y = a, power;
        for(int i = 0;;i++){
            now = k % 2;
            System.out.print(i + "		" + now + "		");
            power = (x * y) % n;
            if(now == 1) {
                System.out.print(x + "x" + y + "(3=)" + power + "		");
                x = power;
            }
            else System.out.print(x + "		");
            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("\		\		" + x + "x" + y + "(3=)" + power);
        return power;
    }

    public static String xTimeShow(String data, String times){
        System.out.println("************************************************************");
        System.out.println("****NOTICE:[Copy lines form second to the last you used]****");
        System.out.println("************************************************************");
        int tar = hex2Dec(data);
        int n = hex2Dec(times);
        int answer = 0;
        for(int i = 0; i < 8; i++){
            if(n%(int)Math.pow(2,i+1) == (int)Math.pow(2,i)){
                answer ^= tar;
                n -= (int)Math.pow(2,i);
            }
            System.out.println(data + "x" + dec2Hex((int)Math.pow(2,i)) + "=" + dec2Hex(tar));
            tar *= 2;
            if(tar >= 256){
                tar -= 256;
                tar ^= 27;
            }
        }
        System.out.println(data + "x" + times + "= ... = ... =" + dec2Hex(answer));
        return dec2Hex(answer);
    }

    private static int hex2Dec(String hex){
        hex = hex.toLowerCase();
        int dec = 0;
        for(int i = 0; i < hex.length(); i++){
            char letter = hex.charAt(i);
            if(letter >= 48 && letter <= 57)dec += (letter - 48) * Math.pow(16,hex.length() - i - 1);
            else dec += (letter - 87) * Math.pow(16,hex.length() - i - 1);
        }
        return dec;
    }

    private static String dec2Hex(int dec){
        String hex = "";
        int[] num = new int[2];
        num[0] = (dec % 256) / 16;
        num[1] = dec % 16;
        if(num[0] >= 0 && num[0] <= 9)hex += num[0];
        else hex += (char)(num[0] + 87);
        if(num[1] >= 0 && num[1] <= 9)hex += num[1];
        else hex += (char)(num[1] + 87);
        return hex;
    }

    public static int[] ECCADD(int a, int p, int xp, int yp, int xq, int yq){
        if(xq == 0 && yq == 0){
            System.out.println("(" + xp + "," + yp + ")");
            return new int[]{xp,yp};
        }
        if(xp == 0 && yp == 0){
            System.out.println("(" + xq + "," + yq + ")");
            return new int[]{xq,yq};
        }
        if(xp == xq && (yp + yq) % p == 0){
            System.out.println("O  (0,0)");
            return new int[]{0,0};
        }
        int lamb;
        if(xp == xq && yp == yq){
            lamb = ((3 * xp * xp + a) * modInverse(2*yp,p)) % p;
        }
        else {
            lamb = ((yq - yp) * modInverse(xq-xp,p)) % p;
        }
        if(lamb < 0) lamb += p;
        System.out.print("lamb = " + lamb);
        int x = (lamb * lamb - xp - xq) % p;
        if(x < 0) x += p;
        int y = (lamb * (xp-x) - yp) % p;
        if(y < 0) y += p;
        System.out.println("   newPoint = (" + x + "," + y + ")");
        return new int[]{x,y};
    }

    public static int[] ECCMUL(int a, int p, int times, int xp, int yp){
        int[] answer = new int[]{xp,yp};
        for(int i = 0; i < times - 1; i++){
            System.out.print((i+2)+"G	");
            answer = ECCADD(a,p,xp,yp,answer[0],answer[1]);
        }
        return answer;
    }

    public static int[][] getECCPoint(int a, int b, int p){
        int[][] point = new int[p+1+(int)(2*Math.sqrt(p))][2];
        point[0] = new int[]{0,0};
        System.out.print("(0,0)");
        int index = 1;
        int[] y2should = new int[p];
        for(int i = 0; i < p; i++){
            y2should[i] = (i*i*i + a * i + b) % p;
        }
        int[] y2real = new int[p];
        for(int i = 0; i < p; i++){
            y2real[i] = (i*i) % p;
        }
        for(int i = 0; i < p; i++){
            for(int j = 0; j < p; j++){
                if(y2should[i] == y2real[j]){
                    point[index][0] = i;
                    point[index][1] = j;
                    System.out.print((",(" + i + "," + j + ")"));
                    index++;
                }
            }
        }
        System.out.println();
        return point;
    }

    public static void main(String[] args) {
        //modInverseShow(10915,31846);//模逆运算
        //modPowerShow(35,27,79);//模幂运算
        //System.out.println(dec2Hex(14));//十进制转十六进制
        //System.out.println(hex2Dec("fb"));//十六进制转十进制
        //xTimeShow("3a","b8");//xTime算法
        //LFSR lfsr = new LFSR(8,"00101");//LFSR画状态图
        //ECCADD(1,11,5,2,2,7);//椭圆曲线加法
        //ECCMUL(1,11,17,2,7);//椭圆曲线乘法
        //getECCPoint(1,6,11);//获得椭圆曲线上的点
    }
}
原文地址:https://www.cnblogs.com/blueflameashe/p/13175421.html