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2013
11-12

POJ 3039 Close Encounter [解题报告] Java

Close Encounter

问题描述 :

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.

输入:

* Line 1: Two positive space-separated integers N and D (1 <= N < D <= 32,767), respectively the numerator and denominator of the given fraction

输出:

* Line 1: Two space-separated integers, respectively the numerator and denominator of the smallest, closest fraction different from the input fraction.

样例输入:

2 3

样例输出:

21845 32767

温馨提示:

INPUT DETAILS:

2/3

OUTPUT DETAILS:

21845/32767 = .666676839503…. ~ 0.666666…. = 2/3.

解题代码:

/* @author: */
import java.util.Scanner;
import java.util.Arrays;
public class Main{

 public static void main(String args[])
{
 Scanner sc=new Scanner(System.in);
 double a,b,e,min;
 int c,d,u=0,v=0;
 a=sc.nextDouble();
 b=sc.nextDouble();
 a=a/b;
 c=d=1;
 min=32768;
 while(c< 32768&&d<32768)
 {
    e=c*1.0/d*1.0;
    if(e==a){
	d++;
	continue;
    }
    if(e>a){
	if(e-a< min){
	 min=e-a;
	 u=c;
	 v=d;
	}
	d++;
    }
   else {
      if(a-e< min){
	 min=a-e;
	 u=c;
	 v=d;
      }
      c++;
   }
  }
  System.out.printf("%d %d",u,v);
  }
}

  1. for(int i=1; i<=m; i++){
    for(int j=1; j<=n; j++){
    dp = dp [j-1] + 1;
    if(s1.charAt(i-1) == s3.charAt(i+j-1))
    dp = dp[i-1] + 1;
    if(s2.charAt(j-1) == s3.charAt(i+j-1))
    dp = Math.max(dp [j - 1] + 1, dp );
    }
    }
    这里的代码似乎有点问题? dp(i)(j) = dp(i)(j-1) + 1;这个例子System.out.println(ils.isInterleave("aa","dbbca", "aadbbcb"));返回的应该是false

  2. 给你一组数据吧:29 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 1000。此时的数据量还是很小的,耗时却不短。这种方法确实可以,当然或许还有其他的优化方案,但是优化只能针对某些数据,不太可能在所有情况下都能在可接受的时间内求解出答案。

  3. #include <cstdio>
    #include <cstring>

    const int MAXSIZE=256;
    //char store[MAXSIZE];
    char str1[MAXSIZE];
    /*
    void init(char *store) {
    int i;
    store['A']=’V', store['B']=’W',store['C']=’X',store['D']=’Y',store['E']=’Z';
    for(i=’F';i<=’Z';++i) store =i-5;
    }
    */
    int main() {
    //freopen("input.txt","r",stdin);
    //init(store);
    char *p;
    while(fgets(str1,MAXSIZE,stdin) && strcmp(str1,"STARTn")==0) {
    if(p=fgets(str1,MAXSIZE,stdin)) {
    for(;*p;++p) {
    //*p=store[*p]
    if(*p<’A’ || *p>’Z') continue;
    if(*p>’E') *p=*p-5;
    else *p=*p+21;
    }
    printf("%s",str1);
    }
    fgets(str1,MAXSIZE,stdin);
    }
    return 0;
    }