hdu4726

Kia's Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 588    Accepted Submission(s): 158

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

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

 

 

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int a[30],b[30];
char A[2000000],B[2000000];
int numa[2000000],numb[1000000];
int cnt[2000000];
int main()
{
    int t,num=1,i,len,j,x,y,min,count,k;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s%s",A,B);
        len=strlen(A);
        for(i=0; i<len; i++)
        {
            numa[i]=A[i]-'0';
            numb[i]=B[i]-'0';
        }
        if(len==1)
        {
            printf("Case #%d: %d ",num++,(numa[0]+numb[0])%10);
            continue;
        }
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(i=0; i<len; i++)
        {
            a[numa[i]]++;
            b[numb[i]]++;
        }
        for(k=0; k<len;)
        {
            min=-1;
            for(i=0; i<=9; i++)
            {
                if(k==0&&i==0)
                    continue;
                for(j=0; j<=9; j++)
                {
                    if(k==0&&j==0)
                        continue;
                    if(a[i]&&b[j]&&(i+j)%10>min)
                    {
                        x=i;
                        y=j;
                        min=(x+y)%10;
                    }
                }
            }
            a[x]--;
            b[y]--;
            cnt[k++]=min;
            if(k>1)
            {
                while(a[x]>0&&b[y]>0)        //剪枝，一旦找到比较大的，可以看看是否符合下一位
                {
                    cnt[k++]=min;
                    a[x]--;
                    b[y]--;
                }
            }
        }
        printf("Case #%d: ",num++);
        for(i=0;i<len;i++)
        {
            if(cnt[i]!=0)
            break;
        }
        if(i==len)
        {
            printf("0 ");
            continue;
        }
        for(;i<len;i++)
        printf("%d",cnt[i]);
        printf(" ");
    }
    return 0;
}