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
题目要求是给定两个数，能加合成另一个数，要求这个数最大，然而第一位不能有0加合成。
然而题目给的AB两个数长度达到10^6（一开始这个条件理解错了，以为是AB上界是10^6，然后果断暴力超时了）
不过，虽然AB长度到达10^6，但是每一位毕竟是由0到9数字构成的，而且题目的加合运算是每位进行的。可以考虑统计0到9的个数然后进行贪心。
由于考虑到第一位不能有0加合，对第一位加合情况进行枚举求最大的。
然后就是对后面的位数进行贪心了：
对于不超过10的情况，自然是从9开始贪心，然后8、7、6……
而且由于对于加合成k的情况，每个能加合成k的对都是不同的，自然互不影响，所以对于每一个能加合成k的情况，就把所有的对全部用完，直到A组和B组中有一个减到0。
对于需要模10的情况。如果同样是模10加合成k，那么肯定先考虑模10得到k的，才会去考虑不模得到k-1的，所以考虑完不模的情况就马上考虑模的情况。
此外这题还有注意点，就是A和B中有一组只有0的情况或两组都是只有0的情况，需要特判。
代码：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
int a[15], b[15], ans[1000010];
void Input()
{
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
char ch;
for (;;)
{
ch = getchar();
if (ch == ‘\n‘)
break;
a[ch-‘0‘]++;
}
for (;;)
{
ch = getchar();
if (ch == ‘\n‘)
break;
b[ch-‘0‘]++;
}
}
void Work()
{
int cnt = 0, maxOne = -1, iOne, jOne;
for (int i = 1; i < 10; ++i)
{
if (a[i] == 0)
continue;
for (int j = 1; j < 10; ++j)
{
if (b[j] == 0)
continue;
if (maxOne < (i+j)%10)
{
maxOne = (i+j)%10;
iOne = i;
jOne = j;
}
}
}
if (maxOne != -1)//第一位出现0加一个数的情况
{
ans[cnt] = maxOne;
cnt++;
a[iOne]--;
b[jOne]--;
}
for (int k = 9; k >= 0; k--)
{
for (int i = k; i >= 0; --i)
{
while (a[i] && b[k-i])
{
ans[cnt] = k;
cnt++;
a[i]--;
b[k-i]--;
}
}
for (int i = k; i < 10; ++i)
{
while (a[i] && b[k+10-i])
{
ans[cnt] = k;
cnt++;
a[i]--;
b[k+10-i]--;
}
}
}
int i = 0;
while (ans[i] == 0 && i < cnt)//排除第一位出现0的情况
i++;
if (i == cnt)//所有位都是0的情况
{
printf("0\n");
return;
}
for (; i < cnt; ++i)
printf("%d", ans[i]);
printf("\n");
}
int main()
{
//freopen("test.in", "r", stdin);
int T;
scanf("%d", &T);
getchar();
for (int times = 1; times <= T; ++times)
{
printf("Case #%d: ", times);
Input();
Work();
}
return 0;
}
ACM学习历程—HDU 4726 Kia's Calculation（ 贪心&&计数排序）