# 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

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#define LL long long
using namespace std;
const int N = 1000005;
char str1[N],str2[N];
int num1,num2;
int res[N];
int main()
{
int tcase,t = 1;
scanf("%d",&tcase);
while(tcase--)
{

scanf("%s%s",str1,str2);
if(strcmp(str1,"0")==0){
printf("Case #%d: ",t++);
printf("%s\n",str2);
continue;
}
if(strcmp(str2,"0")==0){
printf("Case #%d: ",t++);
printf("%s\n",str1);
continue;
}
int len = strlen(str1);
memset(num1,0,sizeof(num1));
memset(num2,0,sizeof(num2));
for(int i=0; i<len; i++)
{
num1[str1[i]-'0']++;
num2[str2[i]-'0']++;
}
int high = -1,x,y;
for(int i=1; i<=9; i++)
{
for(int j=1; j<=9; j++)
{
if(num1[i]&&num2[j]&&high<(i+j)%10)
{
x = i;
y = j;
high = (i+j)%10;
}
}
}
num1[x]--;
num2[y]--;
int cnt = 0,zero = 0;
res[cnt++] = high;
if(high==0) zero++;
printf("Case #%d: ",t++);
if(zero){
printf("0\n");
continue;
}
for(int l=1; 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=9;i>=0&&flag;i--){
for(int j=0;j<=9&&flag;j++){
if(i-j<0&&num1[j]&&num2[i-j+10]){
num1[j]--;
num2[i-j+10]--;
res[cnt++] = i;
flag = false;
}else if(i-j>=0&&num1[j]&&num2[i-j]){
num1[j]--;
num2[i-j]--;
res[cnt++] = i;
flag = false;
}
}
}

}
for(int i=0;i<cnt;i++){
printf("%d",res[i]);
}
printf("\n");
}
return 0;
}

posted @ 2016-09-27 21:08  樱花庄的龙之介大人  阅读(...)  评论(...编辑  收藏