HDU4726-Kia's Calculation

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Kia's Calculation

                                                                             Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                                        Total Submission(s): 3508    Accepted Submission(s): 743


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
解题思路: 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 <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <cmath>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <functional>#include <climits>using namespace std;#define LL long longconst int INF=0x3f3f3f3f;int a[20],b[20],t,cas;char s1[1000050],s2[1000050],ans[1000050];int main(){    cas=1;    scanf("%d",&t);    while(t--)    {        memset(a,0,sizeof a);        memset(b,0,sizeof b);        scanf("%s%s",s1,s2);        int len=strlen(s1);        if(len==1)        {            printf("Case #%d: %d\n",cas++,(s1[0]-'0'+s2[0]-'0')%10);            continue;        }        for(int i=0;s1[i];i++) a[s1[i]-'0']++;        for(int i=0;s2[i];i++) b[s2[i]-'0']++;        for(int i=0;i<len;i++)        {            int flag=0;            for(int j=9;j>=0&&!flag;j--)            {                for(int k=0;k<=9;k++)                {                    if(j-k>=0)                    {                        if(i==0&&(k==0||j-k==0)) continue;                        else                        {                            if(a[k]>0&&b[j-k]>0)                            {                                ans[i]=j+'0';                                a[k]--;                                b[j-k]--;                                flag=1;                                break;                            }                        }                    }                    else                    {                        if(a[k]>0&&b[j+10-k]>0)                        {                            ans[i]=j+'0';                            a[k]--;                            b[j+10-k]--;                            flag=1;                            break;                        }                    }                }            }        }        ans[len]='\0';        printf("Case #%d: ",cas++);        int i=0;        while(ans[i]=='0'&&i<len-1) i++;        while(i<len)            printf("%c",ans[i++]);        printf("\n");    }}

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