UVA 10069 ---Distinct Subsequences +DP+大数

可以定义dp[i][j]表示第一个串的前i个字符中含有第二个串的前j个字符的总情况数;

则:如dp[i][j]=dp[i-1][j],如果str1[i]==str2[j]则dp[i][j]+=dp[i-1][j-1];

初始时讲所有的dp[i][0]赋值为1,其他为0。

然后这个题目需要用到大数,可以用C++重载运算符,或者是java的大数类;

我用的是java,第一次用java的大数,感觉还不错 :)

代码如下:

import java.util.*;
import java.math.*;
public class Main {
	public static void main(String[] args){

		int t;
		String str1;
		String str2;
		BigInteger dp[][] =new BigInteger[10010][120];
		Scanner cin = new Scanner(System.in);
	    t=cin.nextInt();
	    for(int ii=1;ii<=t;ii++){
	        str1=cin.next();
	        str2=cin.next();
	        int len1 =str1.length();
	        int len2 =str2.length();
	        for(int i=0;i<=len1;i++)
	        	for(int j=0;j<=len2;j++)
	        		 dp[i][j]=BigInteger.ZERO;
	        for(int i=0;i<=len1;i++)
	        	dp[i][0]=BigInteger.ONE;
	        for(int i=1;i<=len1;i++)
	        	for(int j=1;j<=len2;j++){

	        		dp[i][j]=dp[i-1][j];
	        		if(str1.charAt(i-1)==str2.charAt(j-1))
	        			dp[i][j]=dp[i][j].add(dp[i-1][j-1]);
	        	}
	        System.out.println(dp[len1][len2]);
	    }
	}

}
时间: 2024-10-11 02:04:32

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