K - Kia's Calculation (贪心）

Kia's Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3902    Accepted Submission(s): 784

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

1 5958 3036

 

 

Sample Output

Case #1: 8984

 

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

 

 

Recommend

zhuyuanchen520

/*
* @Author: lyuc
* @Date:   2017-04-28 19:34:17
* @Last Modified by:   lyuc
* @Last Modified time: 2017-04-28 20:19:20
*/

/*题意：定义一种加法法则，两个整数相加的时候不用进位，现在给你两个位数相等的整数，用着每个整数任意组合，让你构造出
 *    两个数的和最大。
 *
 *思路：先统计每个数中的0-9个出现了多少次，然后最高位特殊处理，剩下的都构造最大的数
 */
#include <bits/stdc++.h>
#define MAXN 1000005
using namespace std;
int t;
char s1[MAXN],s2[MAXN];
int num[MAXN];//用来存放数字
int vis[2][10];
void init(){
    memset(vis,0,sizeof vis);
    memset(num,'\0',sizeof num);
}
int main(){
    // freopen("in.txt","r",stdin);
    scanf("%d",&t);
    for(int ca=1;ca<=t;ca++){
        printf("Case #%d: ",ca);
        init();
        scanf("%s%s",s1,s2);
        int n=strlen(s1);
        if(n==1){//如果只有一位的话，直接输出就可以了
            printf("%d\n",(s1[0]-'0'+s2[0]-'0')%10);
            continue;
        }
        //统计每个数字出现的次数
        for(int i=0;i<n;i++){
            vis[0][s1[i]-'0']++;
            vis[1][s2[i]-'0']++;
        }
        for(int res=0;res<n;res++){//从最高位开始枚举
            for(int i=9;i>=0;i--){//i枚举的是给最后的和的第res位的数，因为要尽量找大的嘛，所以从9开始枚举
                bool flag=false;
                for(int j=0;j<=9;j++){
                    if(vis[0][j]==0) continue;//如果第一个数中没有出现的数字那么根本不需要考虑了
                    int ok=(i+10-j)%10;//ok表示和j(第一个数中出现的数字)组成i的在第二个数中出现的数字
                    if(res==0&&(j==0||ok==0)) continue;//如果是res==0(当前枚举的是最高位，那么就不能出现前导零)
                    if(vis[1][ok]){//如果ok在第二个数中还有那么就可以构造了
                        vis[0][j]--;
                        vis[1][ok]--;
                        num[res]=i;
                        flag=true;
                        break;
                    }
                }
                if(flag==true) break;
            }
        }
        int res=0;
        while(res<n&&num[res]==0) res++;//去前导零
        if(res==n){
            puts("0");
        }else{
            for(;res<n;res++){
                printf("%d",num[res]);
            }
            printf("\n");
        }
    }
    return 0;
}