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

题目要求是给定两个数，能加合成另一个数，要求这个数最大，然而第一位不能有0加合成。

然而题目给的AB两个数长度达到10^6（一开始这个条件理解错了，以为是AB上界是10^6，然后果断暴力超时了）

不过，虽然AB长度到达10^6，但是每一位毕竟是由0到9数字构成的，而且题目的加合运算是每位进行的。可以考虑统计0到9的个数然后进行贪心。

由于考虑到第一位不能有0加合，对第一位加合情况进行枚举求最大的。

然后就是对后面的位数进行贪心了：

对于不超过10的情况，自然是从9开始贪心，然后8、7、6……

而且由于对于加合成k的情况，每个能加合成k的对都是不同的，自然互不影响，所以对于每一个能加合成k的情况，就把所有的对全部用完，直到A组和B组中有一个减到0。

对于需要模10的情况。如果同样是模10加合成k，那么肯定先考虑模10得到k的，才会去考虑不模得到k-1的，所以考虑完不模的情况就马上考虑模的情况。

此外这题还有注意点，就是A和B中有一组只有0的情况或两组都是只有0的情况，需要特判。

代码：

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           using namespace std;int a[15], b[15], ans[1000010];void Input(){ memset(a, 0, sizeof(a)); memset(b, 0, sizeof(b)); char ch; for (;;) { ch = getchar(); if (ch == '\n') break; a[ch-'0']++; } for (;;) { ch = getchar(); if (ch == '\n') break; b[ch-'0']++; }}void Work(){ int cnt = 0, maxOne = -1, iOne, jOne; for (int i = 1; i < 10; ++i) { if (a[i] == 0) continue; for (int j = 1; j < 10; ++j) { if (b[j] == 0) continue; if (maxOne < (i+j)%10) { maxOne = (i+j)%10; iOne = i; jOne = j; } } } if (maxOne != -1)//第一位出现0加一个数的情况 { ans[cnt] = maxOne; cnt++; a[iOne]--; b[jOne]--; } for (int k = 9; k >= 0; k--) { for (int i = k; i >= 0; --i) { while (a[i] && b[k-i]) { ans[cnt] = k; cnt++; a[i]--; b[k-i]--; } } for (int i = k; i < 10; ++i) { while (a[i] && b[k+10-i]) { ans[cnt] = k; cnt++; a[i]--; b[k+10-i]--; } } } int i = 0; while (ans[i] == 0 && i < cnt)//排除第一位出现0的情况 i++; if (i == cnt)//所有位都是0的情况 { printf("0\n"); return; } for (; i < cnt; ++i) printf("%d", ans[i]); printf("\n");}int main(){ //freopen("test.in", "r", stdin); int T; scanf("%d", &T); getchar(); for (int times = 1; times <= T; ++times) { printf("Case #%d: ", times); Input(); Work(); } return 0;}