BZOJ 1684: [Usaco2005 Oct]Close Encounter
题目
1684: [Usaco2005 Oct]Close Encounter
Time Limit: 5 Sec Memory Limit: 64 MBDescription
Lacking even a fifth grade education, the cows are having trouble with a fraction problem from their textbook. Please help them. The problem is simple: Given a properly reduced fraction (i.e., the greatest common divisor of the numerator and denominator is 1, so the fraction cannot be further reduced) find the smallest properly reduced fraction with numerator and denominator in the range 1..32,767 that is closest (but not equal) to the given fraction. 找一个分数它最接近给出一个分数. 你要找的分数的值的范围在1..32767
Input
* Line 1: Two positive space-separated integers N and D (1 <= N < D <= 32,767), respectively the numerator and denominator of the given fraction
Output
* Line 1: Two space-separated integers, respectively the numerator and denominator of the smallest, closest fraction different from the input fraction.
Sample Input
Sample Output
OUTPUT DETAILS:
21845/32767 = .666676839503.... ~ 0.666666.... = 2/3.
HINT
Source
题解
这道题就是看你细不细心= =就是数据类型转换的问题，还有必须要用long double~
代码
1 /*Author:WNJXYK*/ 2 #include<cstdio> 3 using namespace std; 4 int n,m; 5 long double tmp; 6 int ansb,ansa; 7 long double delta=1e30; 8 inline long double abs(long double x){ 9 if (x<0) return -x; 10 return x; 11 } 12 int main(){ 13 scanf("%d%d",&n,&m); 14 tmp=(long double)n/(long double)m; 15 for (int b=1;b<=32767;b++){ 16 int fz=((long double)n/(long double)m*(long double)b); 17 if (abs((long double)(fz-1)/(long double)b-(long double)n/(long double)m)<delta && (fz-1)*m!=n*b){ 18 delta=abs((long double)(fz-1)/(long double)b-(long double)n/(long double)m); 19 ansa=fz-1; 20 ansb=b; 21 } 22 if (abs((long double)(fz)/(long double)b-(long double)n/(long double)m)<delta && fz*m!=n*b){ 23 delta=abs((long double)(fz)/(long double)b-(long double)n/(long double)m); 24 ansa=fz; 25 ansb=b; 26 } 27 if (abs((long double)(fz+1)/(long double)b-(long double)n/(long double)m)<delta && (1+fz)*m!=n*b){ 28 delta=abs((long double)(fz+1)/(long double)b-(long double)n/(long double)m); 29 ansa=fz+1; 30 ansb=b; 31 } 32 } 33 printf("%d %d\n",ansa,ansb); 34 return 0; 35 }
