how to integrals with fractions

Test instructionsThe classic Egyptian score problem, which is to give a true fraction, is to use the smallest number of unit fractions to represent this score. If there are multiple scenarios, make each score as large as possible, that is, the denominator as small as possible. There will be K prohibited unit fractions.Analysis:ida* algorithm. When in the order of the denominator increment, if the current considered score is 1/e, the remaining maxd-d+1

Rocky Valley 1458 Order of fractionsAddress: http://www.luogu.org/problem/show?pid=1458Title DescriptionEnter a natural number N, for a minimal fraction of A/b (numerator and denominator coprime fractions), to meet 1Here is an example, when n=5, all solutions are:0/1 1/5 1/4 1/3 2/5 1/2 3/5 2/3 3/4 4/5 1/1Given a natural number n,1Note: The greatest common divisor of ①0 and any natural number is the natural number ② coprime refers to the two natural n

Title Link: https://projecteuler.net/problem=71If nD and HCF (n,d) =1, it is called a reduced proper fraction.n/d true fractions sorted in ascending order, nearest number to 3/7, dJava Program: Public classp71{voidRun () {calculate (); } voidcalculate () {intMax_n = 1000000; LongA = 3; Longb = 7; LongR = 0; Longs = 1; intQ = 0; Longp = 0; for(q = max_n;q>2;q--) {p= (a*q-1)/b; if(p*s>r*3) {s=Q; R=p; }} System.out.println (R+"/"+s); } Public Sta

10976 Fractions AgainIt's easy-to-see-every fraction in the form 1 K (k > 0), we can always find the positive integers x and y, X ≥y, such that:1 k = 1 x + 1 y now we question Is:can you write a program that counts how many such pairs of x and y th ere is for any given k? Input input contains no more than to lines, each giving a value of K (0 Sample Input212Sample Output21/2 = 1/6 + 1/31/2 = 1/4 + 1/481/12 = 1/156 + 1/131/12 = 1/84 + 1/141/12 = 1/60 +

Topic Portal1 /*2 x>=y, 1/x 3 */4#include 5#include 6#include 7#include 8#include 9#include string>Ten#include One#include Set> A#include - using namespacestd; - the Const intMAXN = 1e4 +Ten; - Const intINF =0x3f3f3f3f; - structNODE - { + intx, y; - }NODE[MAXN]; + A intMainvoid)//UVA 10976 Fractions again?! at { - //freopen ("uva_10976.in", "R", stdin); - - intK; - while(SCANF ("%d", k) = =1) - { in intCNT =0; -

Tags: des style blog color Io OS ar JavaOctal Fractions Time limit:1000 ms Memory limit:10000 K Total submissions:6669 Accepted:3641 DescriptionFractions in octal (base 8) notation can be expressed exactly in decimal notation. for example, 0.75 in octal is 0.953125 (7/8 + 5/64) in decimal. all Octal numbers of N digits to the right of the octal point can be expressed in no more than 3N decimal digits to the right

question: ask for such a four natural number p,q,r,s (pAnalysis: The original type of the same points, simplified finishing to get: 2Using System;namespace consoleapplication1{ class program { static void Main (string[] args) { int p, q, R, s, n = 0; for (P = 2, p Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. The sum of----fractions of C # fun Prog

Topic Links:id=1131 "target=" _blank ">click here~Although Java simulates water over, I see someone's wonderful code, posted and shared with you.Import java.math.*;import java.util.*;class main{public static void Main (String args[]) {Scanner cin = new Scanner (System. in); BigDecimal Eight = new BigDecimal (8), while (Cin.hasnext ()) {String num;num = Cin.nextline (); BigDecimal ans = new BigDecimal (0); BigDecimal tmp = new BigDecimal (1), for (int i = 2;i The algorithm is to convert division

1,2,3...9 These nine numbers make up a fraction with a value of exactly 1/3, how does the group method work?The following program implements this function, please fill in the missing code in the underlined section.#include void Test (int x[]){int a = x[0]*1000 + x[1]*100 + x[2]*10 + x[3];int b = x[4]*10000 + x[5]*1000 + x[6]*100 + x[7]*10 + x[8];if (a*3==b) printf ("%d/%d\n", A, b);}void f (int x[], int k){int i,t;if (k>=9) {Test (x);Return}for (i=k; i{t=x[k]; x[k]=x[i]; x[i]=t;}f (x,k+1);______

Console.Write ("Please enter the number of classes:"); intn =Convert.ToInt32 (Console.ReadLine ()); ArrayList Fenshu=NewArrayList (); Doublesum =0; for(inti =0; I //to enter the elements of a collection with a For loop{Console.Write ("Please enter section"+ (i+1)+"Personal Score:"); Fenshu. ADD (Double. Parse (Console.ReadLine ()));//cast receive data type is double typeSum + =Double. Parse (Fenshu[i]. ToString ());//casts a collection element to a string type and then casts the string type to

One, test instructions:ask for an octal decimal in decimal. Two, ideas:Violent array simulation calculation, pay attention to do not take the decimal division operation1, for octal decimals, converted to decimal, written form analysis:2, simulate the division process:3, output. Three, step:1,0.756[8], ((6/8 + 5)/8 + 7)/8 [ten]2, Division process: such as 0.75I,5/8 Results if the whole number, then 5000/8=625 (5 plus a few 0 can be divisible by 8 to add a few 0)ii,625 can not directly add 7, shou

arithmetic questions to output");int J=input.nextint ();Random t=new random ();Char arr[] = {' + ', '-', ' * ', '/'};for (int i=0;iSystem.out.print (T.nextint (30));System.out.print (Arr[t.nextint (4)]);System.out.println (T.nextint (30) +1+ "=");}Break}Default:System.out.println ("The number entered is incorrect, please enter 0 or 1");}}}Third, the experimental resultsFour reasons why you can't make a classFirst, because forget how to use the random number so that all the attention in this, ca

1#include 2 inta[30010],b[30010];3 intMain ()4 {5 inti,j,k,l,m,n,x,y,z;6 while(SCANF ("%d", k) = =1)7 {8n=0;9 for(i=k+1; i2*k;i++)Ten if((k*i)% (i-k) = =0) One { Aa[++n]=k*i/(I-k); -b[n]=i; - } theprintf"%d\n", n); - for(i=1; i) -printf"1/%d = 1/%d + 1/%d\n", K,a[i],b[i]); - } +}Programming is simple, the key is mathematical deduction.∵x≥y∴1/x≤1/y∴1/k=1/x+1/y≤2/ynamely y≤2kand easy to know y>kEnumerate within this range.When valida

#include int main (){int x,y,i,a,j,k,d;for (i=0;i{X=rand ()% (9-1+1) +1;Y=rand ()% (9-1+1) +1;A=rand ()% (4-1+1) +1;J=rand ()% (9-1+1) +1;K=rand ()% (9-1+1) +1;Switch (a){Case 1:if (xprintf ("(%d/%d) * (%d/%d) =\n", x,y,j,k);else if (x>yj>k)printf ("(%d/%d) * (%d/%d) =\n", y,x,k,j);else if (xprintf ("(%d/%d) * (%d/%d) =\n", x,y,k,j);Elseprintf ("(%d/%d) * (%d/%d) =\n", y,x,j,k);Case 2:if (xprintf ("(%d/%d) + (%d/%d) =\n", x,y,j,k);else if (x>yj>k)printf ("(%d/%d) + (%d/%d) =\n", y,x,k,j);else if

;denominator*A.molecule; Temp2= This->denominator*A.denominator; L=gcd (TEMP1,TEMP2); C.molecule=temp1/m; C.denominator=temp2/m; returnC; } Fractionoperator/(Fraction a) {fraction C; intTemp1;///molecule intTemp2;///Denominator intL///Greatest Common Divisortemp1= This->molecule*A.denominator; Temp2= This->denominator*A.molecule; L=gcd (TEMP1,TEMP2); C.molecule=temp1/m; C.denominator=temp2/m; returnC; }Private: intmolecule;///molecule intdenominator;///Denominator}; Fraction:

The method is: a/b+c/d= ((a*d) + (B*C))/(B*D)(1). h file#import @interface Fraction: nsobject //fraction score @property int Numerator,denominator ; //denominator denominator numerator molecule -(void ) print; -(void ) Setto: (int ) n over:(int ) D; -(double ) Converttonum; //convertion convert -(void ) Add: (fraction *) F; -(void ) reduce; //reduce decrease @end(2). m file#import "Fraction.h"@implementation Fraction @synthesize Numerator,denominator;-(void) print{ NSLog(@ "

Problem:Enter a fraction to divide the score into Egyptian fractions.True fraction: fraction of numerator less than denominatorEgypt score: A fraction of a molecule#include Decompose the true number into Egyptian fractions

Main topic:This means entering a k and then finding all the forms that satisfy 1/k = 1/x+1/y, making x>=y.Problem Solving Ideas:After getting this question, see the topic only give x>=y. This condition, feel impossible, but it is not difficult to think carefully, as long as weThe scope of the enumeration is determined first, so we can find the solution that meets our requirements within this interval.Derivation:Because X>=y, so 1/xThen, by the equation 1/k = 1/x+1/y x = (y*k)/(Y-K)In this way, w

#include #include #include #define N JMain (){int a,b,k,i,n,c,d,e,f,j;do{printf ("\ n input Quantity:");scanf ("%d", j);printf ("\ n input type (1. True score 2. Integer)");scanf ("%d", n);Switch (n){Case 1:{for (i=0;i{E=rand ()%100+1;F=rand ()%100+1;K=rand ()%4+1;if (e>=f)printf ("%d/%d", f,e);Elseprintf ("%d/%d", e,f);Switch (k){Case 1:printf ("+");Case 2:printf ("-");Case 3:printf ("*");Case 4:printf ("/");}C=rand ()%100+1;D=rand ()%100+1;if (C>=d)printf ("%d/%d=\t\t", d,c);Elseprintf ("%d/%d

The procedure is as follows#include #include #include void Main (){int A, B, C, d,i,m,n;float p, q;Srand (Unsigned (Time (NULL)));for (i = 0; i {A = rand ()% 100;b = rand ()% 100;c = rand ()% 5;Switch (c){Case 0:printf ("%d +%d =\n", A, b);BreakCase 1:if (A {D = A;A = b;b = D;}printf ("%d-%d =\n", A, b);BreakCase 2:printf ("%d *%d =\n", A, b);BreakCase 3:while (b = = 0){b = rand ()% 100;}printf ("%d/%d =\n", A, b);BreakCase 4:A = rand ()% 100;b = rand ()% 100;m = rand ()% 100;n = rand ()% 100;c