winfab 2199

2199. [HZOI 2016] event vote, 2199 hzoi ★★Input file:hztp.inOutput file:hztp.outSimple comparisonTime Limit: 0.5 s memory limit: 2 MB[Description] There are many and many people in the China-based shopping festival. n students are voting for one event. Now, more than half of the contestant's votes are received and their contestant numbers are requested (random contestant numbers)[Input format] The first lin

1#include 2#include 3#include 4 #defineSC (x) scanf ("%d", (x))5 #definePF (x) printf ("%.4lf\n", X)6 using namespacestd;7 intT;8 DoubleY;9 DoubleRDoublex)Ten { One return(8*pow (x,4) +7*pow (x,3) +2*pow (x,2) +3*x +6); A } - DoubleFind () - { the DoubleFront =0, rear = -, mid; - while((Rear-front) > 1e-6) - { -Mid = (front + rear)/2; + if(R (Mid) Y) -Front = mid + 1e-7; + Else ARear = mid-1e-7; at } - return(front + rear)/2; - } - intMain () - { - SC (T)

Topic linksproblem DescriptionNow,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 = = Y,can you find its solution between 0 and 100;Now try your lucky.InputThe first line of the input contains an integer T (1OutputFor each test case, you should just output one real number (accurate up to 4 decimal places), and which is the solution of the E Quation,or "No solution!", if there is No solution for the equation between 0 and 100.Sample Input2100-4Sample Output1.6152No solution! The two-point sear

Test instructionsFor Y value, find the solution of polynomial 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 = = y between 0 and 100.IdeasFrom 0 to 100, the polynomial is monotonous, so it is solved by the dichotomy method.CodeDoubleCalcDoublex) { return 8*x*x*x*x+7*x*x*x+2*x*x+3*x+6;}intMain () {intT; CIN>>T; while(t--){ DoubleY; CIN>>Y; DoubleL,r; L=0.0, r=100.0; if(Calc (L) >y | | Y>Calc (R)) {cout"No solution!"Endl; }Else{ Doublemid; while((r-l) > (1e-Ten) ) {Mid= (l+r)/2; if(Calc (m

Can you solve this equation?Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 11180 Accepted Submission (s): 5151Problem Descriptionnow,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 = = Y,can you find its solution between 0 and 100 ;Now try your lucky.Inputthe first line of the input contains an integer T (1Outputfor Each test case, you should just output one real number (accurate up to 4 decimal places), which is the solution of The Equati

A bad mood, a simple two-point#include HDU 2199 Can solve this equation

Http://acm.hdu.edu.cn/showproblem.php? Pid = 1, 2199 /* Binary search; */ #include Http://acm.hdu.edu.cn/showproblem.php? Pid = 1, 2899 /* The function is F (x) = 6 * x ^ 7 + 8 * x ^ 6 + 7 * x ^ 3 + 5 * x ^ 2-y * x; Then its function is f' (x) = 42 * x ^ 6 + 48 * x ^ 5 + 21 * x ^ 2 + 10 * x-y; Design the function and calculate the value of the function G' (x) = 42 * x ^ 6 + 48 * x ^ 5 + 21 * x ^ 2 + 10 * x, subtract the input y value; It can be

Link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 2199Can you solve this equation? Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)Total submission (s): 8595 accepted submission (s): 3957Problem descriptionnow, given the equation 8 * x ^ 4 + 7 * x ^ 3 + 2 * x ^ 2 + 3 * x + 6 = Y, can you find its solution between 0 and 100; Now please try your lucky. Inputthe first line of the input contains an integer T (1 Outputfor each test case, You shocould just output on

("%d",N)) + { A for(intk=1; k) at { -scanf"%LF",m); - - DoubleI=0.0; - if(M0)|| M>f ( -)) - { inprintf"No solution!\n");Continue; - } to DoubleFr=0.0, ed=100.0; + for(; Fabs (F (FR)-F (ed)) >1e-4;) - { theI= (fr+ed)/2; * if(f (i) i; $ ElseEd=i;Panax Notoginseng } - { theprintf"%.4lf\n", i); + } A } the } +

Problem Description:now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 = = Y,can you find its solution between 0 and 10 0;Now try your lucky. Input:the first line of the input contains an integer T (1Test instructions: Enter the value of Y, the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 = = y solution, if there is a solution, then the solution must be in [0, 100] between, Output X, if no solution then output no solution!.#include #includeintMain () {DoubleA, B, C, Y, sum, D; intT, X; scanf ("%

and Upper_bound. The return value is a pair of iterator I and J, where I is the first location where path can be placed, and J is the last position where path can be placed. Can be performed: each element in [I,j] is equivalent to path, and [I, J] is[Left,right]One of the largest sub-ranges that conform to the above properties. The algorithm Lower_bound returns the first iterator of the range, the algorithm Upper_bound returns the Past-the-end iterator of the range, and the algorithm Equal_rang

Can you solve this equation? Time limit:2000/1000ms (java/other) Memory limit:32768/32768k (Java/other) total submission (s): Accepted Submis Sion (s): 17Problem descriptionnow,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 = = Y,can you find its solution Betwe En 0 and 100;Now try your lucky.Inputthe first line of the input contains an integer T (1Outputfor Each test case, you should just output one real number (accurate up to 4 decimal places), which is the solution of The Equation,or "No

The key is to determine the scope of the solution and the basis for judging two ticksThe scope of the solution has been determined, because it is an increment function, and the binary basis is compared to the size of Y.#include"Cstdio"#include"CString"#include"algorithm"#include"Cmath"#defineMAXN 505using namespacestd;DoubleCalDoublex) { Doubletemp=8.0*pow (x,4)+7.0*pow (x,3)+2.0*pow (x,2)+3.0*x+6.0; returntemp;}intMain () {intT; scanf ("%d",T); while(t--) { DoubleAns,y; scanf ("%LF",y);

Can you solve this equation?Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 11044 Accepted Submission (s): 5083Problem Descriptionnow,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 = = Y,can you find its solution between 0 and 100 ;Now try your lucky.Inputthe first line of the input contains an integer T (1Outputfor Each test case, you should just output one real number (accurate up to 4 decimal places), which is the solution of The Equat

The main idea is that both ends are approximated, and finally the result is 0MS#include #include#include#include#includeusing namespacestd;Const DoubleEP = 1e-6;DoubleDealDoubleN) { return 8*pow (N,4) +7*pow (N,3) +2*pow (N,2) +3*n +6;}intMain () {intT,i; Doubley; scanf ("%d",t); while(t--) {scanf ("%LF",y); DoubleLo =0, hi = -; if(Y0) || Y>deal ( -) {printf ("No solution!\n"); Continue; } Else { Doublemid; while(hi-lo>EP) {Mid= (Lo+hi)/2; if(Deal (mid)

Returns the root of the equation using the bipartite method. /** Hdu-2199 can you solve the equatino * Mike-W * 2012-4-17 **/# include

Problem Descriptionnow,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 = = Y,can you find its solution between 0 and 100 ;Now try your lucky. Inputthe first line of the input contains an integer T (1#include Double ans (double A)//calculation (8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6){return (8*a*a*a*a+7*a*a*a+2*a*a+3*a+6.0);}Double ans (double);//Declaration of functionint main (){Double L,r,y,mid;int N,flag;scanf ("%d", n);while (n--){scanf ("%lf", y);//l is the left end and R is the right end;l=0;f

Miyu original, post Please note: Reprinted from __________ White House Question address:Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 2199Description:Can you solveThisEquation? Time limit:2000/1000MS (Java/Others) memory limit:32768/32768K

There is a counter based on different parentheses. During the traversal, if the counter is smaller than 0, false is returned. If the counter is not zero after the traversal, false is returned. Import Java. util. secret; public class bracketsmatch

Can You solve this equation? Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 13018 Accepted Submission (s): 5835Problem Description Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 = = Y,can