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C# BigInteger.Normalize方法代码示例

本文整理汇总了C#中Mono.Math.BigInteger.Normalize方法的典型用法代码示例。如果您正苦于以下问题:C# BigInteger.Normalize方法的具体用法?C# BigInteger.Normalize怎么用?C# BigInteger.Normalize使用的例子?那么恭喜您, 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Mono.Math.BigInteger的用法示例。


在下文中一共展示了BigInteger.Normalize方法的15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: BarrettReduction

            public void BarrettReduction(BigInteger x)
            {
                var n = mod;
                uint k = n.length,
                    kPlusOne = k + 1,
                    kMinusOne = k - 1;

                // x < mod, so nothing to do.
                if (x.length < k) return;

                BigInteger q3;

                //
                // Validate pointers
                //
                if (x.data.Length < x.length) throw new IndexOutOfRangeException("x out of range");

                // q1 = x / b^ (k-1)
                // q2 = q1 * constant
                // q3 = q2 / b^ (k+1), Needs to be accessed with an offset of kPlusOne

                // TODO: We should the method in HAC p 604 to do this (14.45)
                q3 = new BigInteger(Sign.Positive, x.length - kMinusOne + constant.length);
                Kernel.Multiply(x.data, kMinusOne, x.length - kMinusOne, constant.data, 0, constant.length, q3.data, 0);

                // r1 = x mod b^ (k+1)
                // i.e. keep the lowest (k+1) words

                var lengthToCopy = x.length > kPlusOne ? kPlusOne : x.length;

                x.length = lengthToCopy;
                x.Normalize();

                // r2 = (q3 * n) mod b^ (k+1)
                // partial multiplication of q3 and n

                var r2 = new BigInteger(Sign.Positive, kPlusOne);
                Kernel.MultiplyMod2p32pmod(q3.data, (int) kPlusOne, (int) q3.length - (int) kPlusOne, n.data, 0,
                    (int) n.length, r2.data, 0, (int) kPlusOne);

                r2.Normalize();

                if (r2 <= x)
                {
                    Kernel.MinusEq(x, r2);
                }
                else
                {
                    var val = new BigInteger(Sign.Positive, kPlusOne + 1);
                    val.data[kPlusOne] = 0x00000001;

                    Kernel.MinusEq(val, r2);
                    Kernel.PlusEq(x, val);
                }

                while (x >= n)
                    Kernel.MinusEq(x, n);
            }
开发者ID:claudiuslollarius,项目名称:monotorrent,代码行数:58,代码来源:BigInteger.cs

示例2: MultiplyByDword

			public static BigInteger MultiplyByDword (BigInteger n, uint f)
			{
				BigInteger ret = new BigInteger (Sign.Positive, n.length + 1);

				uint i = 0;
				ulong c = 0;

				do {
					c += (ulong)n.data [i] * (ulong)f;
					ret.data [i] = (uint)c;
					c >>= 32;
				} while (++i < n.length);
				ret.data [i] = (uint)c;
				ret.Normalize ();
				return ret;

			}
开发者ID:runefs,项目名称:Marvin,代码行数:17,代码来源:BigInteger.cs

示例3: RightShift

			public static BigInteger RightShift (BigInteger bi, int n)
			{
				if (n == 0) return new BigInteger (bi);

				int w = n >> 5;
				int s = n & ((1 << 5) - 1);

				BigInteger ret = new BigInteger (Sign.Positive, bi.length - (uint)w + 1);
				uint l = (uint)ret.data.Length - 1;

				if (s != 0) {

					uint x, carry = 0;

					while (l-- > 0) {
						x = bi.data [l + w];
						ret.data [l] = (x >> n) | carry;
						carry = x << (32 - n);
					}
				} else {
					while (l-- > 0)
						ret.data [l] = bi.data [l + w];

				}
				ret.Normalize ();
				return ret;
			}
开发者ID:runefs,项目名称:Marvin,代码行数:27,代码来源:BigInteger.cs

示例4: LeftShift

			public static BigInteger LeftShift (BigInteger bi, int n)
			{
				if (n == 0) return new BigInteger (bi, bi.length + 1);

				int w = n >> 5;
				n &= ((1 << 5) - 1);

				BigInteger ret = new BigInteger (Sign.Positive, bi.length + 1 + (uint)w);

				uint i = 0, l = bi.length;
				if (n != 0) {
					uint x, carry = 0;
					while (i < l) {
						x = bi.data [i];
						ret.data [i + w] = (x << n) | carry;
						carry = x >> (32 - n);
						i++;
					}
					ret.data [i + w] = carry;
				} else {
					while (i < l) {
						ret.data [i + w] = bi.data [i];
						i++;
					}
				}

				ret.Normalize ();
				return ret;
			}
开发者ID:runefs,项目名称:Marvin,代码行数:29,代码来源:BigInteger.cs

示例5: multiByteDivide

			public static BigInteger [] multiByteDivide (BigInteger bi1, BigInteger bi2)
			{
				if (Kernel.Compare (bi1, bi2) == Sign.Negative)
					return new BigInteger [2] { 0, new BigInteger (bi1) };

				bi1.Normalize (); bi2.Normalize ();

				if (bi2.length == 1)
					return DwordDivMod (bi1, bi2.data [0]);

				uint remainderLen = bi1.length + 1;
				int divisorLen = (int)bi2.length + 1;

				uint mask = 0x80000000;
				uint val = bi2.data [bi2.length - 1];
				int shift = 0;
				int resultPos = (int)bi1.length - (int)bi2.length;

				while (mask != 0 && (val & mask) == 0) {
					shift++; mask >>= 1;
				}

				BigInteger quot = new BigInteger (Sign.Positive, bi1.length - bi2.length + 1);
				BigInteger rem = (bi1 << shift);

				uint [] remainder = rem.data;

				bi2 = bi2 << shift;

				int j = (int)(remainderLen - bi2.length);
				int pos = (int)remainderLen - 1;

				uint firstDivisorByte = bi2.data [bi2.length-1];
				ulong secondDivisorByte = bi2.data [bi2.length-2];

				while (j > 0) {
					ulong dividend = ((ulong)remainder [pos] << 32) + (ulong)remainder [pos-1];

					ulong q_hat = dividend / (ulong)firstDivisorByte;
					ulong r_hat = dividend % (ulong)firstDivisorByte;

					do {

						if (q_hat == 0x100000000 ||
							(q_hat * secondDivisorByte) > ((r_hat << 32) + remainder [pos-2])) {
							q_hat--;
							r_hat += (ulong)firstDivisorByte;

							if (r_hat < 0x100000000)
								continue;
						}
						break;
					} while (true);

					//
					// At this point, q_hat is either exact, or one too large
					// (more likely to be exact) so, we attempt to multiply the
					// divisor by q_hat, if we get a borrow, we just subtract
					// one from q_hat and add the divisor back.
					//

					uint t;
					uint dPos = 0;
					int nPos = pos - divisorLen + 1;
					ulong mc = 0;
					uint uint_q_hat = (uint)q_hat;
					do {
						mc += (ulong)bi2.data [dPos] * (ulong)uint_q_hat;
						t = remainder [nPos];
						remainder [nPos] -= (uint)mc;
						mc >>= 32;
						if (remainder [nPos] > t) mc++;
						dPos++; nPos++;
					} while (dPos < divisorLen);

					nPos = pos - divisorLen + 1;
					dPos = 0;

					// Overestimate
					if (mc != 0) {
						uint_q_hat--;
						ulong sum = 0;

						do {
							sum = ((ulong)remainder [nPos]) + ((ulong)bi2.data [dPos]) + sum;
							remainder [nPos] = (uint)sum;
							sum >>= 32;
							dPos++; nPos++;
						} while (dPos < divisorLen);

					}

					quot.data [resultPos--] = (uint)uint_q_hat;

					pos--;
					j--;
				}

				quot.Normalize ();
				rem.Normalize ();
//.........这里部分代码省略.........
开发者ID:runefs,项目名称:Marvin,代码行数:101,代码来源:BigInteger.cs

示例6: DwordDivMod

			public static BigInteger [] DwordDivMod (BigInteger n, uint d)
			{
				BigInteger ret = new BigInteger (Sign.Positive , n.length);

				ulong r = 0;
				uint i = n.length;

				while (i-- > 0) {
					r <<= 32;
					r |= n.data [i];
					ret.data [i] = (uint)(r / d);
					r %= d;
				}
				ret.Normalize ();

				BigInteger rem = (uint)r;

				return new BigInteger [] {ret, rem};
			}
开发者ID:runefs,项目名称:Marvin,代码行数:19,代码来源:BigInteger.cs

示例7: SingleByteDivideInPlace

			/// <summary>
			/// Performs n / d and n % d in one operation.
			/// </summary>
			/// <param name="n">A BigInteger, upon exit this will hold n / d</param>
			/// <param name="d">The divisor</param>
			/// <returns>n % d</returns>
			public static uint SingleByteDivideInPlace (BigInteger n, uint d)
			{
				ulong r = 0;
				uint i = n.length;

				while (i-- > 0) {
					r <<= 32;
					r |= n.data [i];
					n.data [i] = (uint)(r / d);
					r %= d;
				}
				n.Normalize ();

				return (uint)r;
			}
开发者ID:runefs,项目名称:Marvin,代码行数:21,代码来源:BigInteger.cs

示例8: OddPow

			private BigInteger OddPow (uint b, BigInteger exp)
			{
				exp.Normalize ();
				uint [] wkspace = new uint [mod.length << 1 + 1];

				BigInteger resultNum = Montgomery.ToMont ((BigInteger)b, this.mod);
				resultNum = new BigInteger (resultNum, mod.length << 1 +1);

				uint mPrime = Montgomery.Inverse (mod.data [0]);

				uint pos = (uint)exp.bitCount () - 2;

				//
				// We know that the first itr will make the val b
				//

				do {
					//
					// r = r ^ 2 % m
					//
					Kernel.SquarePositive(resultNum, ref wkspace);
					resultNum = Montgomery.Reduce(resultNum, mod, mPrime);

					if (exp.testBit(pos)) {

						//
						// r = r * b % m
						//

						uint u = 0;

						uint i = 0;
						ulong mc = 0;

						do {
							mc += (ulong)resultNum.data[u + i] * (ulong)b;
							resultNum.data[u + i] = (uint)mc;
							mc >>= 32;
						} while (++i < resultNum.length);

						if (resultNum.length < mod.length) {
							if (mc != 0) {
								resultNum.data[u + i] = (uint)mc;
								resultNum.length++;
								while (resultNum >= mod)
									Kernel.MinusEq(resultNum, mod);
							}
						} else if (mc != 0) {

							//
							// First, we estimate the quotient by dividing
							// the first part of each of the numbers. Then
							// we correct this, if necessary, with a subtraction.
							//

							uint cc = (uint)mc;

							// We would rather have this estimate overshoot,
							// so we add one to the divisor
							uint divEstimate = (uint)((((ulong)cc << 32) | (ulong)resultNum.data[u + i - 1]) /
								(mod.data[mod.length - 1] + 1));

							uint t;

							i = 0;
							mc = 0;
							do {
								mc += (ulong)mod.data[i] * (ulong)divEstimate;
								t = resultNum.data[u + i];
								resultNum.data[u + i] -= (uint)mc;
								mc >>= 32;
								if (resultNum.data[u + i] > t) mc++;
								i++;
							} while (i < resultNum.length);
							cc -= (uint)mc;

							if (cc != 0) {

								uint sc = 0, j = 0;
								uint[] s = mod.data;
								do {
									uint a = s[j];
									if (((a += sc) < sc) | ((resultNum.data[u + j] -= a) > ~a)) sc = 1;
									else sc = 0;
									j++;
								} while (j < resultNum.length);
								cc -= sc;
							}
							while (resultNum >= mod)
								Kernel.MinusEq(resultNum, mod);
						} else {
							while (resultNum >= mod)
								Kernel.MinusEq(resultNum, mod);
						}
					}
				} while (pos-- > 0);

				resultNum = Montgomery.Reduce (resultNum, mod, mPrime);
				return resultNum;

//.........这里部分代码省略.........
开发者ID:Balamir,项目名称:DotNetOpenAuth,代码行数:101,代码来源:BigInteger.cs

示例9: Subtract

			public static BigInteger Subtract (BigInteger big, BigInteger small)
			{
				BigInteger result = new BigInteger (Sign.Positive, big.length);

				uint [] r = result.data, b = big.data, s = small.data;
				uint i = 0, c = 0;

				do {

					uint x = s [i];
					if (((x += c) < c) | ((r [i] = b [i] - x) > ~x))
						c = 1;
					else
						c = 0;

				} while (++i < small.length);

				if (i == big.length) goto fixup;

				if (c == 1) {
					do
						r [i] = b [i] - 1;
					while (b [i++] == 0 && i < big.length);

					if (i == big.length) goto fixup;
				}

				do
					r [i] = b [i];
				while (++i < big.length);

				fixup:

					result.Normalize ();
				return result;
			}
开发者ID:runefs,项目名称:Marvin,代码行数:36,代码来源:BigInteger.cs

示例10: AddSameSign

			/// <summary>
			/// Adds two numbers with the same sign.
			/// </summary>
			/// <param name="bi1">A BigInteger</param>
			/// <param name="bi2">A BigInteger</param>
			/// <returns>bi1 + bi2</returns>
			public static BigInteger AddSameSign (BigInteger bi1, BigInteger bi2)
			{
				uint [] x, y;
				uint yMax, xMax, i = 0;

				// x should be bigger
				if (bi1.length < bi2.length) {
					x = bi2.data;
					xMax = bi2.length;
					y = bi1.data;
					yMax = bi1.length;
				} else {
					x = bi1.data;
					xMax = bi1.length;
					y = bi2.data;
					yMax = bi2.length;
				}
				
				BigInteger result = new BigInteger (Sign.Positive, xMax + 1);

				uint [] r = result.data;

				ulong sum = 0;

				// Add common parts of both numbers
				do {
					sum = ((ulong)x [i]) + ((ulong)y [i]) + sum;
					r [i] = (uint)sum;
					sum >>= 32;
				} while (++i < yMax);

				// Copy remainder of longer number while carry propagation is required
				bool carry = (sum != 0);

				if (carry) {

					if (i < xMax) {
						do
							carry = ((r [i] = x [i] + 1) == 0);
						while (++i < xMax && carry);
					}

					if (carry) {
						r [i] = 1;
						result.length = ++i;
						return result;
					}
				}

				// Copy the rest
				if (i < xMax) {
					do
						r [i] = x [i];
					while (++i < xMax);
				}

				result.Normalize ();
				return result;
			}
开发者ID:runefs,项目名称:Marvin,代码行数:65,代码来源:BigInteger.cs

示例11: ToMont

			public static BigInteger ToMont (BigInteger n, BigInteger m)
			{
				n.Normalize (); m.Normalize ();

				n <<= (int)m.length * 32;
				n %= m;
				return n;
			}
开发者ID:runefs,项目名称:Marvin,代码行数:8,代码来源:BigInteger.cs

示例12: EvenModTwoPow

/* known to be buggy in some cases */
#if false
			private unsafe BigInteger EvenModTwoPow (BigInteger exp)
			{
				exp.Normalize ();
				uint [] wkspace = new uint [mod.length << 1 + 1];

				BigInteger resultNum = new BigInteger (2, mod.length << 1 +1);

				uint value = exp.data [exp.length - 1];
				uint mask = 0x80000000;

				// Find the first bit of the exponent
				while ((value & mask) == 0)
					mask >>= 1;

				//
				// We know that the first itr will make the val 2,
				// so eat one bit of the exponent
				//
				mask >>= 1;

				uint wPos = exp.length - 1;

				do {
					value = exp.data [wPos];
					do {
						Kernel.SquarePositive (resultNum, ref wkspace);
						if (resultNum.length >= mod.length)
							BarrettReduction (resultNum);

						if ((value & mask) != 0) {
							//
							// resultNum = (resultNum * 2) % mod
							//

							fixed (uint* u = resultNum.data) {
								//
								// Double
								//
								uint* uu = u;
								uint* uuE = u + resultNum.length;
								uint x, carry = 0;
								while (uu < uuE) {
									x = *uu;
									*uu = (x << 1) | carry;
									carry = x >> (32 - 1);
									uu++;
								}

								// subtraction inlined because we know it is square
								if (carry != 0 || resultNum >= mod) {
									uu = u;
									uint c = 0;
									uint [] s = mod.data;
									uint i = 0;
									do {
										uint a = s [i];
										if (((a += c) < c) | ((* (uu++) -= a) > ~a))
											c = 1;
										else
											c = 0;
										i++;
									} while (uu < uuE);
								}
							}
						}
					} while ((mask >>= 1) > 0);
					mask = 0x80000000;
				} while (wPos-- > 0);

				return resultNum;
			}
开发者ID:runefs,项目名称:Marvin,代码行数:73,代码来源:BigInteger.cs

示例13: EvenPow

			private unsafe BigInteger EvenPow (uint b, BigInteger exp)
			{
				exp.Normalize ();
				uint [] wkspace = new uint [mod.length << 1 + 1];
				BigInteger resultNum = new BigInteger ((BigInteger)b, mod.length << 1 + 1);

				uint pos = (uint)exp.BitCount () - 2;

				//
				// We know that the first itr will make the val b
				//

				do {
					//
					// r = r ^ 2 % m
					//
					Kernel.SquarePositive (resultNum, ref wkspace);
					if (!(resultNum.length < mod.length))
						BarrettReduction (resultNum);

					if (exp.TestBit (pos)) {

						//
						// r = r * b % m
						//

						// TODO: Is Unsafe really speeding things up?
						fixed (uint* u = resultNum.data) {

							uint i = 0;
							ulong mc = 0;

							do {
								mc += (ulong)u [i] * (ulong)b;
								u [i] = (uint)mc;
								mc >>= 32;
							} while (++i < resultNum.length);

							if (resultNum.length < mod.length) {
								if (mc != 0) {
									u [i] = (uint)mc;
									resultNum.length++;
									while (resultNum >= mod)
										Kernel.MinusEq (resultNum, mod);
								}
							} else if (mc != 0) {

								//
								// First, we estimate the quotient by dividing
								// the first part of each of the numbers. Then
								// we correct this, if necessary, with a subtraction.
								//

								uint cc = (uint)mc;

								// We would rather have this estimate overshoot,
								// so we add one to the divisor
								uint divEstimate = (uint) ((((ulong)cc << 32) | (ulong) u [i -1]) /
									(mod.data [mod.length-1] + 1));

								uint t;

								i = 0;
								mc = 0;
								do {
									mc += (ulong)mod.data [i] * (ulong)divEstimate;
									t = u [i];
									u [i] -= (uint)mc;
									mc >>= 32;
									if (u [i] > t) mc++;
									i++;
								} while (i < resultNum.length);
								cc -= (uint)mc;

								if (cc != 0) {

									uint sc = 0, j = 0;
									uint [] s = mod.data;
									do {
										uint a = s [j];
										if (((a += sc) < sc) | ((u [j] -= a) > ~a)) sc = 1;
										else sc = 0;
										j++;
									} while (j < resultNum.length);
									cc -= sc;
								}
								while (resultNum >= mod)
									Kernel.MinusEq (resultNum, mod);
							} else {
								while (resultNum >= mod)
									Kernel.MinusEq (resultNum, mod);
							}
						}
					}
				} while (pos-- > 0);

				return resultNum;
			}
开发者ID:runefs,项目名称:Marvin,代码行数:98,代码来源:BigInteger.cs

示例14: DwordDiv

			public static BigInteger DwordDiv(BigInteger n, uint d)
			{
				var ret = new BigInteger(Sign.Positive, n.length);

				ulong r = 0;
				uint i = n.length;

				while (i-- > 0)
				{
					r <<= 32;
					r |= n.data[i];
					ret.data[i] = (uint) (r/d);
					r %= d;
				}
				ret.Normalize();

				return ret;
			}
开发者ID:rajkosto,项目名称:DayZeroLauncher,代码行数:18,代码来源:BigInteger.cs

示例15: BigInteger

		public static BigInteger operator * (BigInteger bi1, BigInteger bi2)
		{
			if (bi1 == 0 || bi2 == 0) return 0;

			//
			// Validate pointers
			//
			if (bi1.data.Length < bi1.length) throw new IndexOutOfRangeException ("bi1 out of range");
			if (bi2.data.Length < bi2.length) throw new IndexOutOfRangeException ("bi2 out of range");

			BigInteger ret = new BigInteger (Sign.Positive, bi1.length + bi2.length);

			Kernel.Multiply (bi1.data, 0, bi1.length, bi2.data, 0, bi2.length, ret.data, 0);

			ret.Normalize ();
			return ret;
		}
开发者ID:runefs,项目名称:Marvin,代码行数:17,代码来源:BigInteger.cs


注:本文中的Mono.Math.BigInteger.Normalize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。