HDU 4726 Kia's Calculation

最新推荐文章于 2019-01-08 08:31:31 发布

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本文探讨了如何通过重新排列两个整数的数字来获得最大可能的和，并提供了实现这一目标的算法。通过实例演示了从给定的整数中找到最佳组合的方法。

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

随机组合用不进制来求出最大值

#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;

char a[1000005],b[1000005];
int aa[15],bb[15],ans[1000005];

int main()
{
    int cas,l,k,j,i,v,temp1,temp2,o,temp3;
    scanf("%d",&cas);
    for(int tt=1;tt<=cas;tt++)
    {
        scanf("%s %s",a,b);
        if(!strcmp(a,"0"))
            {printf("Case #%d: ",tt);puts(b);continue;}
        if(!strcmp(b,"0"))
            {printf("Case #%d: ",tt);puts(a);continue;}
        l=strlen(a);
        memset(aa,0,sizeof aa);
        memset(bb,0,sizeof bb);
        for(i=0; i<l; i++)
        {
            aa[a[i]-'0']++;
            bb[b[i]-'0']++;
        }
        v=0;
        int maxx=-1;
        for(j=1; j<10; j++)
        {
            for(k=1; k<10; k++)
            {
                if(aa[j]&&bb[k])
                {
                    if((j+k)%10>maxx)
                    {
                        temp1=j;
                        temp2=k;
                        maxx=(j+k)%10;
                    }
                }
            }
        }
        ans[v++]=maxx;
        aa[temp1]--;
        bb[temp2]--;
        if(maxx==0)
        {
            printf("Case #%d: ",tt);puts("0");continue;
        }
        for(k=9; k>=0; k--)
            for(o=0; o<10; o++)
            {
                for(i=0; i<10; i++)
                {
                    if((o+i)%10==k)
                    {
                        while(aa[o]&&bb[i])
                        {
                            aa[o]--;
                            bb[i]--;
                            ans[v++]=k;
                          //  printf("%d %d %d\n",o,i,k);
                        }
                    }
                }
            }
        printf("Case #%d: ",tt);
        for(i=0;i<v;i++)
            printf("%d",ans[i]);
        printf("\n");
    }
    return 0;
}