当前位置: 首页>>代码示例>>C#>>正文


C# BigInteger.bitLength方法代码示例

本文整理汇总了C#中BigInteger.bitLength方法的典型用法代码示例。如果您正苦于以下问题:C# BigInteger.bitLength方法的具体用法?C# BigInteger.bitLength怎么用?C# BigInteger.bitLength使用的例子?那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在BigInteger的用法示例。


在下文中一共展示了BigInteger.bitLength方法的5个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于系统推荐出更棒的C#代码示例。

示例1: squareAndMultiply

 internal static BigInteger squareAndMultiply(BigInteger x2, BigInteger a2, BigInteger exponent,BigInteger modulus, int n2  )
 {
     BigInteger res = x2;
     for (int i = exponent.bitLength() - 1; i >= 0; i--) {
         res = monPro(res,res,modulus, n2);
         if (BitLevel.testBit(exponent, i)) {
             res = monPro(res, a2, modulus, n2);
         }
     }
     return res;
 }
开发者ID:sailesh341,项目名称:JavApi,代码行数:11,代码来源:Division.cs

示例2: slidingWindow

        /*Implements the Montgomery modular exponentiation based in <i>The sliding windows algorithm and the Mongomery
         *Reduction</i>.
         *@ar.org.fitc.ref "A. Menezes,P. van Oorschot, S. Vanstone - Handbook of Applied Cryptography";
         *@see #oddModPow(BigInteger, BigInteger,
         *                           BigInteger)
         */
        internal static BigInteger slidingWindow(BigInteger x2, BigInteger a2, BigInteger exponent,BigInteger modulus, int n2)
        {
            // fill odd low pows of a2
            BigInteger []pows = new BigInteger[8];
            BigInteger res = x2;
            int lowexp;
            BigInteger x3;
            int acc3;
            pows[0] = a2;

            x3 = monPro(a2,a2,modulus,n2);
            for (int i = 1; i <= 7; i++){
                pows[i] = monPro(pows[i-1],x3,modulus,n2) ;
            }

            for (int i = exponent.bitLength()-1; i>=0;i--){
                if( BitLevel.testBit(exponent,i) ) {
                    lowexp = 1;
                    acc3 = i;

                    for(int j = java.lang.Math.max(i-3,0);j <= i-1 ;j++) {
                        if (BitLevel.testBit(exponent,j)) {
                            if (j<acc3) {
                                acc3 = j;
                                lowexp = (lowexp << (i-j))^1;
                            } else {
                                lowexp = lowexp^(1<<(j-acc3));
                            }
                        }
                    }

                    for(int j = acc3; j <= i; j++) {
                        res = monPro(res,res,modulus,n2);
                    }
                    res = monPro(pows[(lowexp-1)>>1], res, modulus,n2);
                    i = acc3 ;
                }else{
                    res = monPro(res, res, modulus, n2) ;
                }
            }
            return res;
        }
开发者ID:sailesh341,项目名称:JavApi,代码行数:48,代码来源:Division.cs

示例3: nextProbablePrime

        /**
         * It uses the sieve of Eratosthenes to discard several composite numbers in
         * some appropriate range (at the moment {@code [this, this + 1024]}). After
         * this process it applies the Miller-Rabin test to the numbers that were
         * not discarded in the sieve.
         *
         * @see BigInteger#nextProbablePrime()
         * @see #millerRabin(BigInteger, int)
         */
        internal static BigInteger nextProbablePrime(BigInteger n)
        {
            // PRE: n >= 0
            int i, j;
            int certainty;
            int gapSize = 1024; // for searching of the next probable prime number
            int []modules = new int[primes.Length];
            bool[] isDivisible = new bool[gapSize];
            BigInteger startPoint;
            BigInteger probPrime;
            // If n < "last prime of table" searches next prime in the table
            if ((n.numberLength == 1) && (n.digits[0] >= 0)
                    && (n.digits[0] < primes[primes.Length - 1])) {
                for (i = 0; n.digits[0] >= primes[i]; i++) {
                    ;
                }
                return BIprimes[i];
            }
            /*
             * Creates a "N" enough big to hold the next probable prime Note that: N <
             * "next prime" < 2*N
             */
            startPoint = new BigInteger(1, n.numberLength,
                    new int[n.numberLength + 1]);
            java.lang.SystemJ.arraycopy(n.digits, 0, startPoint.digits, 0, n.numberLength);
            // To fix N to the "next odd number"
            if (n.testBit(0)) {
                Elementary.inplaceAdd(startPoint, 2);
            } else {
                startPoint.digits[0] |= 1;
            }
            // To set the improved certainly of Miller-Rabin
            j = startPoint.bitLength();
            for (certainty = 2; j < BITS[certainty]; certainty++) {
                ;
            }
            // To calculate modules: N mod p1, N mod p2, ... for first primes.
            for (i = 0; i < primes.Length; i++) {
                modules[i] = Division.remainder(startPoint, primes[i]) - gapSize;
            }
            while (true) {
                // At this point, all numbers in the gap are initialized as
                // probably primes
                java.util.Arrays<Object>.fill(isDivisible, false);
                // To discard multiples of first primes
                for (i = 0; i < primes.Length; i++) {
                    modules[i] = (modules[i] + gapSize) % primes[i];
                    j = (modules[i] == 0) ? 0 : (primes[i] - modules[i]);
                    for (; j < gapSize; j += primes[i]) {
                        isDivisible[j] = true;
                    }
                }
                // To execute Miller-Rabin for non-divisible numbers by all first
                // primes
                for (j = 0; j < gapSize; j++) {
                    if (!isDivisible[j]) {
                        probPrime = startPoint.copy();
                        Elementary.inplaceAdd(probPrime, j);

                        if (millerRabin(probPrime, certainty)) {
                            return probPrime;
                        }
                    }
                }
                Elementary.inplaceAdd(startPoint, gapSize);
            }
        }
开发者ID:sailesh341,项目名称:JavApi,代码行数:76,代码来源:Primality.cs

示例4: inplaceModPow2

        /**
         * Performs {@code x = x mod (2<sup>n</sup>)}.
         *
         * @param x a positive number, it will store the result.
         * @param n a positive exponent of {@code 2}.
         */
        internal static void inplaceModPow2(BigInteger x, int n)
        {
            // PRE: (x > 0) and (n >= 0)
            int fd = n >> 5;
            int leadingZeros;

            if ((x.numberLength < fd) || (x.bitLength() <= n)) {
                return;
            }
            leadingZeros = 32 - (n & 31);
            x.numberLength = fd + 1;
            x.digits[fd] &= (leadingZeros < 32) ? (java.dotnet.lang.Operator.shiftRightUnsignet(-1, leadingZeros)) : 0;
            x.cutOffLeadingZeroes();
        }
开发者ID:sailesh341,项目名称:JavApi,代码行数:20,代码来源:Division.cs

示例5: isProbablePrime

        /**
         * @see BigInteger#isProbablePrime(int)
         * @see #millerRabin(BigInteger, int)
         * @ar.org.fitc.ref Optimizations: "A. Menezes - Handbook of applied
         *                  Cryptography, Chapter 4".
         */
        internal static bool isProbablePrime(BigInteger n, int certainty)
        {
            // PRE: n >= 0;
            if ((certainty <= 0) || ((n.numberLength == 1) && (n.digits[0] == 2))) {
                return true;
            }
            // To discard all even numbers
            if (!n.testBit(0)) {
                return false;
            }
            // To check if 'n' exists in the table (it fit in 10 bits)
            if ((n.numberLength == 1) && ((n.digits[0] & 0XFFFFFC00) == 0)) {
                return (java.util.Arrays<Object>.binarySearch(primes, n.digits[0]) >= 0);
            }
            // To check if 'n' is divisible by some prime of the table
            for (int i = 1; i < primes.Length; i++) {
                if (Division.remainderArrayByInt(n.digits, n.numberLength,
                        primes[i]) == 0) {
                    return false;
                }
            }
            // To set the number of iterations necessary for Miller-Rabin test
            int iJ;
            int bitLength = n.bitLength();

            for (iJ = 2; bitLength < BITS[iJ]; iJ++) {
                ;
            }
            certainty = java.lang.Math.min(iJ, 1 + ((certainty - 1) >> 1));

            return millerRabin(n, certainty);
        }
开发者ID:sailesh341,项目名称:JavApi,代码行数:38,代码来源:Primality.cs


注:本文中的BigInteger.bitLength方法示例由纯净天空整理自Github/MSDocs等开源代码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。