XTU 1236 fraction

Fraction
accepted:168		submit:1061
Time limit:1000 MS		Memory limit:65536 KB

Fractionproblem Description: Everyone has silly periods, especially for Renshengge. It's A sunny day, no one knows "what happened to Renshengge," Renshengge says that he wants to the change all decimal fractions Between 0 and 1 to fraction. In addtion, he says decimal fractions was too complicate, and set that [Math processing Error] was much more co Nvient than 0.33333 ... As an example-to-support his theory. So, Renshengge lists a lot of numbers in textbooks and starts he great work. To him dissapoint, he soon realizes that the denominator of the fraction is very big which kills the simplicity that s Upport of his theory. But Renshengge was famous for his persistence, so he decided to sacrifice some accuracy of fractions. Ok, in him new solution, he confines the denominator in [1,1000] and figure out the least absolute different fractions wit H The decimal fraction under his restriction. If several fractions satifies the restriction, he chooses the smallest one with simplest formation. Input The first line contains a number T (no more than 10000) which represents the number of test cases. And there followed T lines, each line contains a finite decimal fraction x that satisfies [Math processing ERR or]. Output For each test case, transform X in Renshengge ' s rule. Sample Input 3 0.9999999999999 0.3333333333333 0.2222222222222 Sample Output 1/1 1/3 2/9 Tip You can use a double to save X;

Looks very complicated problem, actually is water question, do not be intimidated by the topic! Since the denominator is 1-1000, each time you enumerate all the denominators, you can select the nearest number. Test instructions: Enter a decimal number to output the nearest score, which must be the simplest fraction. Attached code:

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <cmath>5 using namespacestd;6 intgcdintAintb)7 {8     intc,t;9     if(a<b)Ten     { Onet=a,a=b,b=T; A     } -      while(b) -     { thec=a%b; -A=b; -b=C; -     } +     returnA; - } + intMain () A { at     inti,j,t; -     DoubleS,minn; -scanf"%d",&T); -      while(t--) -     { -scanf"%LF",&s); in         intA=0, b=1; -minn=s; to          for(i=1; i<= +; i++)//the denominator of enumeration 1-1000 +         { -j=s*i+0.5;//Find the Molecule the             Doublef=j*1.0/I;//calculate the result of this fraction *             DoubleP=fabs (f-s);//compare to the original number $             if(minn>p)Panax Notoginseng             { -minn=p; theA=J; +b=i; A             } the         } +         intR=GCD (A, b);//seeking greatest common divisor, simplifying -printf"%d/%d\n", a/r,b/R); $     } $     return 0; -}

XTU 1236 fraction