HDU 4726 Kia's Calculation

Kia's Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 295    Accepted Submission(s): 80

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 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

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

 

Recommend

zhuyuanchen520

 

题意： 有2个合法的整数。 长度为 10^6。 数字的每一位都能移动， 但移动后的整数一定要是合法的， 即无前导零。 使得 A + B 最大

思路： 因为固定搭配不变

例如 要得到5 ，   （0， 5）（1，4）（2，3） （6， 9） （7，8） 互不干扰。

如果就统计 

A中  0~9分别出现多少个，

B中 0~9分别出现多少个。

除了第一位外， 其他位都是要大就大。

第一位不能出现0， 即可。

#include 
#include 
#include 
using namespace std;
const int V = 1000000 + 50;
int a[20], b[20], T, s = 1;
char ch[V], dh[V], ans[V];
int main() {
    int i, j, k;
    scanf("%d", &T);
    while(T--) {
        memset(a, 0, sizeof(a));
        memset(b, 0, sizeof(b));
        scanf("%s%s", &ch, &dh);
        int len = strlen(ch);
        if(len == 1) {
            printf("Case #%d: %d\n", s++, (ch[0] - '0' + dh[0] - '0') % 10);
            continue;
        }
        for(i = 0; ch[i]; ++i)
            a[ch[i] - '0']++;
        for(i = 0; dh[i]; ++i)
            b[dh[i] - '0']++;
        for(i = 0; i < len; ++i) { //结果的每一位
            int flag = 0;
            for(j = 9; j >= 0 && !flag; --j) { //从大到小找
                for(k = 0; k <= 9; ++k) {
                    if(j - k >= 0) {
                        if(i == 0 && (k == 0 || j - k == 0))
                            ;
                        else {
                            if(a[k] > 0 && b[j - k] > 0) {
                                ans[i] = j + '0';
                                a[k]--;
                                b[j - k]--;
                                flag = 1;
                                break;
                            }
                        }
                    }
                    if(i == 0 && (k == 0 || j + 10 - k == 0))
                        ;
                    else {
                        if(a[k] > 0 && b[j + 10 - k] > 0) {
                            ans[i] = j + '0';
                            a[k]--;
                            b[j + 10 - k]--;
                            flag = 1;
                            break;
                        }
                    }
                }
            }
        }
        ans[i] = '\0';
        if(ans[0] == '0')
            printf("Case #%d: 0\n", s++);
        else
            printf("Case #%d: %s\n", s++, ans);
    }
}