HUST信心大涨迎省赛之《我要冲银牌》K——字符串——Kia's Calculation

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

 大意：先判第一个数，再判后面的，从i=9开始输出符合的多少的组数

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAX = 1000;
int t1[MAX][MAX],t2[MAX][MAX];
int x[MAX],y[MAX];
int main()
{
    //freopen ("K.in", "r", stdin);
    int T;
    scanf("%d",&T);
    int xy = 1;
    while(T--){
        memset(t1,0,sizeof(t1));
        memset(t2,0,sizeof(t2));
        int n,m;
        scanf("%d%d",&n,&m);
        int temp1 = 1;
       while(n){
          x[temp1++] = n%10;
          n/=10;
       }
       int temp2 = 1;
       while(m){
         y[temp2++] = m%10;
         m/=10;
       }
       sort(x+1,x+temp1);
       sort(y+1,y+temp2);
       int x0 = 1;
       temp1 --;
       temp2 --;
       do{
           for(int i = 1; i <=  temp1;i++){
               t1[x0][i] = x[i];
           }
           x0++;
       }
       while(next_permutation(x+1,x+temp1+1));
        int y0 = 1;
        do{
        for(int i = 1; i <= temp2; i++)
              t2[y0][i] = y[i];
        y0++;
       }while(next_permutation(y+1,y+temp2+1));

       //printf ("x0: %d y0: %d
", x0, y0);
           int sum1 = 0;
           for(int i = 1; i <= x0;i++){
            for(int j = 1; j <= y0;j++){
            int sum = 0;
              if(t1[i][1] == 0 || t2[j][1] == 0)
                  continue;
              int k = 1;
            int temp3 = temp1;
            int temp4 = temp2;
            if (temp3 > temp4)
            {
                while(temp3 > temp4){
                sum = sum*10 + t1[i][k];
                    temp4++;
                   k++;
                }
            }
            if (temp4 > temp3)
            {
                while(temp4 > temp3){
                sum = sum*10 + t2[j][k];
                  temp3++;
                  k++;
                }
            }
            if(temp1 > temp2){
           for(int m1 = k,m2 = 1; m1 <=temp3;m1++,m2++)
            sum = (t1[i][m1] + t2[j][m2])%10 + sum * 10;
            sum1 = max(sum1,sum);
            }
            else {
                for(int m1 = 1,m2 = k; m2 <= temp4; m1++,m2++)
                    sum = (t1[i][m1] + t2[j][m2])%10 + sum *10;
                sum1 = max(sum1,sum);
            }
            }
         }
        printf("Case #%d:",xy++);
        printf(" %d
",sum1);
        }
        return 0;
 }