HDU 4726 Kia's Calculation

题目链接：

https://www.nitacm.com/problem_show.php?pid=4600

题目：

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

题意：

给你两个非常大的数（长度小于1e6）,你可以对这两个数上的任意位置进行交换，没有交换次数的限制，然后让你求出这两个数能够组成的最大的和是多少，这里的加法和一般的加法有区别，for exmple:99999+11111=00000，对应的位置为和再 mod10。

题解：

用贪心的思想，先求出两个相加里面有几个9，几个8，几个7。。。。。。最后值得注意的就是前导0的处理。

代码：

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define met(a,b) memset(a,b,sizeof(a)) #define inf 0x3f3f3f3f const int maxn = 1e6+1000; char s1[maxn],s2[maxn]; int num1[10+10],num2[10+10]; int num[maxn]; int main() { int t; int id=1; scanf("%d",&t); while(t--) { scanf("%s%s",s1,s2); int len=strlen(s1); met(num,0); met(num1,0); met(num2,0); for(int i=0; i<len;i++) num1[s1[i]-'0']++; for(int i=0;i<len;i++) num2[s2[i]-'0']++; printf("Case #%d: ",id++); if(len==1) { printf("%d\n",(s1[0]-'0'+s2[0]-'0')); continue; } int MAX=-inf; int pos_x=0; int pos_y=0; for(int i=1;i<=9;i++) for(int j=1;j<=9;j++) { if(num1[i]&&num2[j]&&(i+j)>MAX) { pos_x=i; pos_y=j; MAX=(i+j); } } num1[pos_x]--; num2[pos_y]--; int length=0; num[length++]=MAX; for(int i=9;i>=0;i--) { for(int j=0;j<=9;j++) { if(num1[j]) { if(j<=i) { int k=i-j; int cnt=min(num1[j],num2[k]); num1[j]-=cnt; num2[k]-=cnt; while(cnt--) num[length++]=i; } else { int k=10+i-j; if(k>9) continue; int cnt=min(num1[j],num2[k]); num1[j]-=cnt; num2[k]-=cnt; while(cnt--) num[length++]=i; } } } } int k=0; for(k=0;k<length-1;k++) { if(num[k]!=0) { break; } } for(int i=k;i<length;i++) printf("%d",num[i]); printf("\n"); } }