precision 5520

There are two containers C1 and C2. Initially, C1 has a number of values of V, giving you K operations. Each time we repeat these K operations N times, finally, we will ask you what the number is in C2.N 1: The number of loop operations is huge. It is a question of matrix concatenation.2: K operations can generate a matrix. n k operations are the N power of the matrix.3: multiply it by the initial matrix.It is not difficult to construct the matrix, that is, if else writes for half a day.Finally,

Header files: #include The function atof () is used to convert a string to a double-precision floating-point number (double) whose prototype is:Double atof (const char* str);The name of Atof () is derived from the abbreviation of the ASCII to floating point numbers, which scans the parameter str string, skips the preceding whitespace characters (such as spaces, tab indents, etc., which can be detected by the Isspace () function), The conversion is not

product of two parameters*/public static double Mul (Double v1,double v2) {BigDecimal B1 = new BigDecimal (double.tostring (v1));BigDecimal b2 = new BigDecimal (double.tostring (v2));Return b1.multiply (B2). Doublevalue ();}/*** Provide (relative) accurate division operation, when the occurrence of an endless situation, accurate to* After the decimal point 10 digits, the later numbers are rounded.* @param v1 Dividend* @param v2 Divisor* @return two parameters of the quotient*/public static doub

HDU 2256 Problem of Precision number theory matrix fast power, hdu2256 You must obtain the integer (√ 2 + √ 3) of 2n and then mod 1024. (√ 2 + √ 3) 2n = (5 + 2 √ 6) n If the value is calculated directly, double is used to store the value. When n is large, the loss of precision will increase and the expected result cannot be obtained. We found that (5 + 2 √ 6) n + (5-2 √ 6) n is an integer (the even power o

Many combination problems will be used to Cattleya number, the growth rate is fast, should write a high-precision respect for a bit ~1#include 2#include 3#include 4 #defineML 5495 using namespacestd;6 intkt[ the][550];7 intlen[ the];8 intGetlen (intOrd)9 {Ten intpos; one for(inti=ml;i>=0; i--) a if(kt[ord][i]!=0) - { -pos=i; the break; - } - returnpos+1; - } + voidMutintOrd) - { + for(intI=0; i1];

Link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4382 There are two containers C1 and C2. Initially, C1 has a number of values of V, giving you k operations. Each time we repeat these K operations n times, finally, we will ask you what the number is in C2. N 1: The number of loop operations is huge. It is a question of matrix concatenation. 2: K operations can generate a matrix. n k operations are the N power of the matrix. 3: multiply it by the initial matrix. It is not difficult to const

fourth point and the distance between the center and the denominator is also 2 (BC-AD), so the process of operation is not to divide operations, as long as the comparison of two molecules can be, high-precision problems can only be solved with BigInteger.ImportJava.util.Scanner;ImportJava.util.*;Importjava.math.*;ImportJava.io.*;classpo{BigInteger x, y;} Public classmain{ Public Static BooleanSolve (po p1, po p2, PO P3, po p) {//---------------------

YK-JJW (JSW) Series Product OverviewYK-JJW Series single-phase precision AC power supply, using the International Advanced Sine Energy Distribution program Power Regulation technology, represents the latest development level of AC voltage regulator technology. The circuit consists of sinusoidal energy distribution and high-power filter. It has the advantages of high overload capacity, good reliability, strong anti-jamming ability, high efficiency, lon

Reason: JS in accordance with 2 binary to deal with decimal subtraction, on the basis of arg1 arg2 precision extension or inverse extension matching, so the following situation occurs.JavaScript (JS) decimal point subtraction problem, is a JS bug such as 0.3*1 = 0.2999999999, the following list can be perfect to find the corresponding accuracy of four JS algorithm function Accdiv (arg1,arg2) {var t1=0,t2=0,r1,r2; Try{t1=arg1.tostring (). Split (".")

/*** Because Java's simple type does not accurately operate on floating-point numbers, this tool class provides accurate floating-point arithmetic, including subtraction and rounding. */ Public classArith {//Default division operation Precision Private Static Final intDef_div_scale = 10;//This class cannot be instantiated PrivateArith () {}/*** provides accurate addition operations. * * @paramv1 * Summand *@paramv2 * Addend *@returntwo para

First of all, thank Rujia for the "Algorithmic Competition Primer".As we all know, the unsigned long long in C + + has an upper limit, and if we want to calculate very large integers, we are overwhelmed, so I wrote a high-precision class that allows large integers to arithmeticThis class uses strings for input and output, and uses arrays for storage and processing, and by simulating arithmetic, the subtraction ratio of large integers can be computed.S

Continue to learn Java high-precision, today is written to seek n!.First, attach the source code:　　ImportJava.util.Scanner;ImportJava.math.BigInteger; Public classMain { Public Static voidMain (String []args) {Scanner cin=NewScanner (system.in); BigInteger A,n,i; while(Cin.hasnext ()) {a=Cin.nextbiginteger (); N=Biginteger.one; for(I=biginteger.one; I.compareto (a) I.add (Biginteger.one)) n=n.multiply (i); SYSTEM.OUT.PRINTLN (n); }

(Double.tostring (v2)); returnb1.subtract (B2). Doublevalue (); }/*** provides accurate multiplication operations. *@paramv1 by multiplier *@paramV2 Multiplier *@returnproduct of two parameters*/ PublicStaticDoubleMulDoubleV1,Doublev2) {BigDecimal B1=NewBigDecimal (double.tostring (v1)); BigDecimal B2=NewBigDecimal (Double.tostring (v2)); returnb1.multiply (B2). Doublevalue (); }/*** Provide (relative) accurate division operations, when there are no more than an endless situation, accurate to *

Take a look at what has changed in the next place. 1, open the Excel worksheet, click the "File" button 2, open the left list of the interface click "Options" 3, pop-up Excel Options window, click the Advanced tab 4, on the right window to navigate to the calculation of this workbook options, check the "Set precision to lock display precision" checkbox 5,

1Add (A, b) {//Add2 varC, D, E;3 Try {4c = a.tostring (). Split (".") [1].length;5}Catch(f) {6c = 0;7 }8 Try {9D = b.tostring (). Split (".") [1].length;Ten}Catch(f) { Oned = 0; A } - returnE = Math.pow (ten, Math.max (c, D)), ( This. Mul (A, E) + This. Mul (b, E))/e; - }, theSub (A, b) {//Subtract - varC, D, E; - Try { -c = a.tostring (). Split (".") [1].length; +}Catch(f) { -c = 0; + } A Try { atD