四川省ACM竞赛（2013）---K - Kia's Calculation

K - Kia's Calculation

Time Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%I64d & %I64u

Submit Status

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 
     
     
     
     

 

ＡＣ　ＣＯＤＥ：

#include <iostream>
#include <cstring>
#define MAX 1000001
using namespace std;
char t1[MAX];
char t2[MAX];
void output(int a,int n)
{
	for(int i=0;i<n;i++)
		cout<<a;
}
int main()
{
	int T;
	int count=0;
	cin>>T;
	while(T--)
	{
		int s1[11]={0};
		int s2[11]={0};
		scanf("%s%s",t1,t2);
		int i,j;
		int len=0;
		for(i=0;t1[i];i++)
		{
			len++;
			s1[ t1[i]-'0' ]++;
		}
		for(i=0;t2[i];i++)
		{
			s2[ t2[i]-'0' ]++;
		}
		cout<<"Case #"<<++count<<": ";
		if(len==1)
		{
			cout<<(t1[0]+t2[0]-'0'-'0')%10<<endl;
			continue;
		}
		int left=1;
		int right=8;
		int t=0;
		int count1,count2;
		for(i=1;i<10;i++)
		{
			if(s1[i]==0)continue;
			for(j=1;j<10;j++)
			{
				if(s2[j]==0)continue;
				if((i+j)%10>t)
				{
					t=(i+j)%10;
					count1=i;
					count2=j;
				}
			}
		}
		if(t==0)
		{	
			cout<<0<<endl;
			continue;
		}
		cout<<t;
		s1[count1]--;
		s2[count2]--;
		int sum[11]={0};
		for(i=9;i>=0;i--)        // 1234 5678
		{
			for(j=0;j<10;j++)
			{
				if(s2[j]==0)continue;
				int t=i-j;
				if(t<0)t+=10;
				if(s1[t]==0)continue;
				if(t==2)
				{
			//		cout<<s2[7]<<endl;
			//		cout<<j<<" "<<i<<endl;
				}
				if(s1[t]>s2[j])
				{
					sum[i] += s2[j];
					s1[t]-=s2[j];
					s2[j]=0;
				}
				else
				{
					sum[i] += s1[t];	
					s2[j]-=s1[t];
					s1[t]=0;
				}
			}
		}
		for(i=9;i>=0;i--)
		{	
			output(i,sum[i]);
		}
		cout<<endl;
	}
	return 0;
}