hdu 4726(贪心)

Kia's Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3291    Accepted Submission(s): 703

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

 

题意:给出两个长度不超过 10^6 的数字串,数字串可以打乱后随机组合,但是打乱重组后不能有前导 0,将这两个数字串相加,相加的规则是每一位相加,不实现进位.问能够得到最大的结果串是多少?

题解:贪心求解,从高位到低位,每次贪心选择相加起来最大的数字,最高位要单独处理,因为最高位要求被加数,加数,结果都不能为 0.然后在贪心的过程中不能去枚举加数和被加数,只能枚举结果,不然会超时. 当位数只有1，加数或者被加数中间有一个为0时单独处理.

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#define LL long long
using namespace std;
const int N = ;
char str1[N],str2[N];
int num1[],num2[];
int res[N];
int main()
{
    int tcase,t = ;
    scanf("%d",&tcase);
    while(tcase--)
    {

        scanf("%s%s",str1,str2);
        if(strcmp(str1,"")==){
            printf("Case #%d: ",t++);
            printf("%s\n",str2);
            continue;
        }
        if(strcmp(str2,"")==){
            printf("Case #%d: ",t++);
            printf("%s\n",str1);
            continue;
        }
        int len = strlen(str1);
        memset(num1,,sizeof(num1));
        memset(num2,,sizeof(num2));
        for(int i=; i<len; i++)
        {
            num1[str1[i]-'']++;
            num2[str2[i]-'']++;
        }
        int high = -,x,y;
        for(int i=; i<=; i++)
        {
            for(int j=; j<=; j++)
            {
                if(num1[i]&&num2[j]&&high<(i+j)%)
                {
                    x = i;
                    y = j;
                    high = (i+j)%;
                }
            }
        }
        num1[x]--;
        num2[y]--;
        int cnt = ,zero = ;
        res[cnt++] = high;
        if(high==) zero++;
        printf("Case #%d: ",t++);
        if(zero){
            printf("0\n");
            continue;
        }
        for(int l=; l<len; l++)
        {
            /*
            TLE
            for(int i=0; i<=9; i++)
            {
                for(int j=0; j<=9; j++)
                {
                    if(num1[i]&&num2[j]&&MAX<(i+j)%10)
                    {
                        x = i;
                        y = j;
                        MAX = (i+j)%10;
                    }
                }
            }*/
            bool flag = true;
            for(int i=;i>=&&flag;i--){
                for(int j=;j<=&&flag;j++){
                    if(i-j<&&num1[j]&&num2[i-j+]){
                        num1[j]--;
                        num2[i-j+]--;
                        res[cnt++] = i;
                        flag = false;
                    }else if(i-j>=&&num1[j]&&num2[i-j]){
                        num1[j]--;
                        num2[i-j]--;
                        res[cnt++] = i;
                        flag = false;
                    }
                }
            }

        }
        for(int i=;i<cnt;i++){
            printf("%d",res[i]);
        }
        printf("\n");
    }
    return ;
}