K

Kia's Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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 10⁶, 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

算法：贪心

题解：根据题目意思来，我们只要将所有的数字出现的个数都记录一下（桶排），然后你就遍历依次相加取最大就行了。要注意的你需要单独判断第一个数，它不能有0，因为你的变化是不能把0变成第一位的，然后你还要注意的是，你的结果不能有前导0。

#include <iostream>
#include <cstdio>
#include <memory.h>

using namespace std;

int visa[15], visb[15];
char ans[1000007];
string a, b;

int main() {
    int T;
    int cas = 0;
    scanf("%d", &T);
    while(T--) {
        for(int i = 0; i < 10; i++) {
            visa[i] = visb[i] = 0;
        }
        cin >> a >> b;
        int lena = a.size();
        int lenb = b.size();
        for(int i = 0; i < lena; i++) {
            visa[a[i] - '0']++;
        }
        for(int i = 0; i < lenb; i++) {
            visb[b[i] - '0']++;
        }
        int posa, posb, maxx = -1;
        for(int i = 0; i < 10; i++) {    //找出第一个数
            for(int j = 0; j < 10; j++) {
                if(i != 0 && j != 0 && visa[i] && visb[j] && maxx < (i + j) % 10) {
                    maxx = (i + j) % 10;
                    posa = i;
                    posb = j;
                }
            }
        }
        int len = 0;
        printf("Case #%d: ", ++cas);
        if(maxx >= 0) {    //如果第一个数存在，则存储下来
            ans[len++] = maxx + '0';
            visa[posa]--;
            visb[posb]--;
        }
        for(int k = 9; k >= 0; k--) {        //寻找之后的数字，每次取最大
            for(int i = 0; i < 10; i++) {
                for(int j = 0; j < 10; j++) {
                    while(visa[i] > 0 && visb[j] > 0 && (i + j) % 10 == k) {
                        visa[i]--;
                        visb[j]--;
                        ans[len++] = k + '0';
                    }
                }
            }
        }
        int mark = 1;
        for(int i = 0; i < len; i++) {    //需要判断前导0
            if(mark && i == len - 1) {
                printf("%c", ans[i]);
            } else if(mark && ans[i] != '0') {
                mark = 0;
                printf("%c", ans[i]);
            } else if(!mark) {
                printf("%c", ans[i]);
            }
        }
        printf("
");
    }
    return 0;
}