HDU 4726 Kia's Calculation （贪心算法）

Kia's Calculation

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 513 Accepted Submission(s): 142

Problem Description

Doctor Ghee is teaching Kia how to calculate the sum of two integers. But Kia is so careless and alway forget to carry a number when the sum of two digits exceeds 9. For example, when she calculates 4567+5789, she will get 9246, and for 1234+9876, she will get 0. Ghee is angry about this, and makes a hard problem for her to solve:
Now Kia has two integers A and B, she can shuffle the digits in each number as she like, but leading zeros are not allowed. That is to say, for A = 11024, she can rearrange the number as 10124, or 41102, or many other, but 02411 is not allowed.
After she shuffles A and B, she will add them together, in her own way. And what will be the maximum possible sum of A "+" B ?

 

 

Input

The rst line has a number T (T <= 25) , indicating the number of test cases.
For each test case there are two lines. First line has the number A, and the second line has the number B.
Both A and B will have same number of digits, which is no larger than 10 ⁶, and without leading zeros.

 

 

Output

For test case X, output "Case #X: " first, then output the maximum possible sum without leading zeros.

 

 

Sample Input

1 5958 3036

 

 

Sample Output

Case #1: 8984

 

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

 

 

题意： 有2个合法的整数。 长度为 10^6。 数字的每一位都能移动， 但移动后的整数一定要是合法的， 即无前导零。 使得 A + B 最大

 

思路：贪心算法

import java.io.*;
import java.util.*;

public class Main {
	BufferedReader bu;
	PrintWriter pw;
	int n;
	int[] a = new int[12];
	int[] b = new int[12];

	public static void main(String[] args) throws IOException {
		new Main().work();
	}

	void work() throws IOException {
		bu = new BufferedReader(new InputStreamReader(System.in));
		pw = new PrintWriter(new OutputStreamWriter(System.out), true);
		n = Integer.parseInt(bu.readLine());
		for (int p = 1; p <= n; p++) {

			String s1 = bu.readLine();
			String s2 = bu.readLine();

			Arrays.fill(a, 0);
			Arrays.fill(b, 0);
			
			for (int i = 0; i < s1.length(); i++) {
				a[s1.charAt(i) - '0']++;
			}

			for (int i = 0; i < s2.length(); i++) {
				b[s2.charAt(i) - '0']++;
			}
			//获取第一个最大的数字
			int t = getFirst();
			pw.print("Case #"+p+": ");
			pw.print(t);
			if (t == 0) {//如果第一个数字为0，则后面的数字，都为0
				pw.println();
				continue;
			}
			// 获取后面的数字
			for (int i = 9; i >= 0; i--) {
				int ans = 0;
				for (int j = 0; j <= 9; j++) {
					if ((i - j >= 0) && a[j] != 0 && b[i - j] != 0) {
						int m = Math.min(a[j], b[i - j]);
						ans += m;
						a[j] -= m;
						b[i - j] -= m;
					}
					if ((10 + i - j <= 9) && a[j] != 0 && b[10 + i - j] != 0) {
						int m = Math.min(a[j], b[10 + i - j]);
						ans += m;
						a[j] -= m;
						b[10 + i - j] -= m;
					}
				}
				for (int j = 1; j <= ans; j++) {
					pw.print(i);
				}
			}
			pw.println();
		}
	}
	//获取第一个数字
	int getFirst() {
		int i, j;
		for (i = 9; i >= 1; i--) {

			for (j = 1; j <= 9; j++) {
				if ((i - j > 0) && a[j] != 0 && b[i - j] != 0) {
					a[j]--;
					b[i - j]--;
					break;
				}
				if ((10 + i - j <= 9) && a[j] != 0 && b[10 + i - j] != 0) {
					a[j]--;
					b[10 + i - j]--;
					break;
				}
			}
			if (j <= 9)
				break;

		}
		return i;
	}
}