HDU 4726 Kia's Calculation

题目链接：

https://www.nitacm.com/problem_show.php?pid=4600

题目：

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

题意：

给你两个非常大的数（长度小于1e6）,你可以对这两个数上的任意位置进行交换，没有交换次数的限制，然后让你求出这两个数能够组成的最大的和是多少，这里的加法和一般的加法有区别，for exmple:99999+11111=00000，对应的位置为和再 mod10。

题解：

用贪心的思想，先求出两个相加里面有几个9，几个8，几个7。。。。。。最后值得注意的就是前导0的处理。

代码：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define met(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3fconst int maxn = 1e6+1000;char s1[maxn],s2[maxn];int num1[10+10],num2[10+10];int num[maxn];int main(){    int t;    int id=1;    scanf("%d",&t);    while(t--)    {        scanf("%s%s",s1,s2);        int len=strlen(s1);        met(num,0);        met(num1,0);        met(num2,0);        for(int i=0; i<len;i++)            num1[s1[i]-'0']++;        for(int i=0;i<len;i++)            num2[s2[i]-'0']++;        printf("Case #%d: ",id++);        if(len==1)        {            printf("%d\n",(s1[0]-'0'+s2[0]-'0')%10);            continue;        }        int MAX=-inf;        int pos_x=0;        int pos_y=0;        for(int i=1;i<=9;i++)            for(int j=1;j<=9;j++)            {                if(num1[i]&&num2[j]&&(i+j)%10>MAX)                {                    pos_x=i;                    pos_y=j;                    MAX=(i+j)%10;                }            }        num1[pos_x]--;        num2[pos_y]--;        int length=0;        num[length++]=MAX;        for(int i=9;i>=0;i--)        {            for(int j=0;j<=9;j++)            {                if(num1[j])                {                    if(j<=i)                    {                        int k=i-j;                        int cnt=min(num1[j],num2[k]);                        num1[j]-=cnt;                        num2[k]-=cnt;                        while(cnt--)                            num[length++]=i;                    }                    else                    {                        int k=10+i-j;                        if(k>9)                            continue;                        int cnt=min(num1[j],num2[k]);                        num1[j]-=cnt;                        num2[k]-=cnt;                        while(cnt--)                            num[length++]=i;                    }                }            }        }        int k=0;        for(k=0;k<length-1;k++)        {            if(num[k]!=0)            {                break;            }        }        for(int i=k;i<length;i++)            printf("%d",num[i]);        printf("\n");    }}