Fraction |
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| Accepted : 51 | Submit : 435 | |
| 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 change all decimal fractions between 0 and 1 to fraction. In addtion, he says decimal fractions are too complicate, So, RenShengGe lists a lot of numbers in textbooks and starts his great work. To his dissapoint, he soon realizes that the denominator of the fraction may be very big which kills the simplicity that support of his theory. But RenShengGe is famous for his persistence, so he decided to sacrifice some accuracy of fractions. Ok, In his new solution, he confines the denominator in [1,1000] and figure out the least absolute different fractions with the decimal fraction under his InputThe 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 . OutputFor each test case, transform x in RenShengGe‘s rule. Sample Input3 0.9999999999999 0.3333333333333 0.2222222222222 Sample Output1/1 1/3 2/9 tipYou can use double to save x; |
题意：把一个小数化成分数，分母不会超过1000。
题解：首先要注意的是如果x<0.001,答案就是0和1/1000最接近x的那个；
先预处理小数，排序，再二分找到最接近x的那个数。
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
struct node {
double x;
int d,f;
} s[1000010];
int len;
int cmp(node a,node b) {
if(a.x==b.x) {
return a.f<b.f;
}
return a.x<b.x;
}
void init() {
len=0;
s[len].x=1.0;
s[len].d=1;
s[len++].f=1;
s[len].x=0.0;
s[len].d=1;
s[len++].f=0;
for(int i=2; i<=1000; i++) {
for(int j=1; j<i; j++) {
double k=j*1.0/(i*1.0);
s[len].x=k;
s[len].d=i;
s[len++].f=j;
}
}
sort(s,s+len,cmp);
}
int gcd(int b,int a) {
return b==0?a:gcd(a%b,b);
}
int main() {
init();
int t;
scanf("%d",&t);
double x;
while(t--) {
scanf("%lf",&x);
int l=0,r=len-1;
int L=0,R=len-1;
int mid=(l+r)>>1;
int id=-1;
while(l<r) {
mid=(l+r)>>1;
if(s[mid].x==x) {
id=mid;
break;
}
if(s[mid].x>x) {
R=r;
r=mid-1;
} else {
L=l;
l=mid+1;
}
}
int F,D;
if(id==-1) {//找最接近
double Min=1;
for(int i=L; i<=R; i++) {
double p=fabs(x-s[i].x);
if(Min>p) {
Min=p;
id=i;
}
}
}
F=s[id].f;
D=s[id].d;
int k=gcd(D,F);
printf("%d/%d\n",F/k,D/k);
}
return 0;
}