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  • 2010-2011 ACM-ICPC, NEERC, Moscow Subregional Contest Problem J. Joke 水题

    Problem J. Joke

    题目连接:

    http://codeforces.com/gym/100714

    Description

    The problem is to cut the largest possible number of circles with diameter y out of a stripe of length x
    and width y.

    Input

    The only line of input consists of two positive real numbers x and y with 9-digit precision separated by
    spaces. The integers may be written without decimal point.

    Output

    Output a single integer — the maximum number of circles one can cut out of the stripe.

    Sample Input

    6.3 0.9

    Sample Output

    7

    Hint

    题意

    给你两个数,问你A/B是多少,保证小数点后9为小数以内。

    题解:

    乘以1e9,然后再除就好了

    代码

    import java.io.*;
    import java.math.*;
    import java.util.*;
    
    public class Main
    {
        public static void main(String argv[]) throws Exception
        {
        	Scanner cin = new Scanner(System.in);
        	String x = cin.next() , y = cin.next();
        	{
        		int find = 0;
        		for(int i = 0 ; i < x.length() ; ++ i) if( x.charAt(i) =='.' ) find = 1;
        		if( find == 0 ) x += '.';
        	}
        	{
        		int find = 0;
        		for(int i = 0 ; i < y.length() ; ++ i) if( y.charAt(i) =='.' ) find = 1;
        		if( find == 0 ) y += '.';
        	}
        	int m1 = 0 , m2 = 0;
        	{
        		int find = 0;
        		for(int i = 0 ; i < x.length() ; ++ i){
        			if( x.charAt(i) == '.' )  find = 1;
        			else if( find == 1 )  ++ m1;
        		}
        	}
        	{
        		int find = 0;
        		for(int i = 0 ; i < y.length() ; ++ i){
        			if( y.charAt(i) == '.' )  find = 1;
        			else if( find == 1 )  ++ m2;
        		}
        	}
        	int ms = Math.max( m1 , m2 );
        	for(int i = m1 ; i < ms ; ++ i) x+='0';
        	for(int i = m2 ; i < ms ; ++ i) y+='0';
        	BigInteger A = BigInteger.ZERO , B = BigInteger.ZERO;
    	    for(int i = 0 ; i < x.length() ; ++ i){
    	    	if( x.charAt(i) != '.' ){
    	    		int add = x.charAt(i) - '0';
    	    		A = A.multiply( BigInteger.valueOf(10) );
    	    		A = A.add( BigInteger.valueOf(add) );
    	    	}
    	    }  
    	    for(int i = 0 ; i < y.length() ; ++ i){
    	    	if( y.charAt(i) != '.' ){
    	    		int add = y.charAt(i) - '0';
    	    		B = B.multiply( BigInteger.valueOf(10) );
    	    		B = B.add( BigInteger.valueOf(add) );
    	    	}
    	    }
    	    System.out.println( A.divide(B) );
        }
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/5752305.html
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