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 ?
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.
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
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;
}
}