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 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

159583036

 

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 <cstdio>#include <cstring>#include <algorithm>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);    }}