hdu 4726 Kia's Calculation(贪心,好题)

题目链接

Kia's Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3283    Accepted Submission(s): 702

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

一道很好的贪心题,不过处处是坑,先确定最高位,不能有前导0,然后依次确定后面的位,做法详见代码。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<vector>using namespace std;const int MAXN=2*1e6+100;char s1[MAXN],s2[MAXN];int ans[MAXN];int cnt1[20],cnt2[20];int main(){int cas;scanf("%d",&cas);for(int k=1;k<=cas;k++){memset(cnt1,0,sizeof(cnt1));memset(cnt2,0,sizeof(cnt2));memset(ans,0,sizeof(ans));scanf("%s%s",s1,s2);int len=strlen(s1);if(len==1){int a=s1[0]-'0',b=s2[0]-'0';printf("Case #%d: %d\n",k,(a+b)%10);continue;}for(int i=0;i<len;i++) cnt1[s1[i]-'0']++,cnt2[s2[i]-'0']++;int p=0,q=0,tt=-1;for(int i=1;i<=9;i++) for(int j=1;j<=9;j++) if(cnt1[i]&&cnt2[j])  { if((i+j)%10>tt) tt=(i+j)%10,p=i,q=j; }ans[1]=tt;int tot=1;cnt1[p]--,cnt2[q]--;for(int l=9;l>=0;l--){for(int i=0;i<=9;i++){if(!cnt1[i]) continue;int p;if(i<=l)p=l-i;else p=l-i+10;int q=min(cnt1[i],cnt2[p]);cnt1[i]-=q;cnt2[p]-=q;while(q--) ans[++tot]=l;}}printf("Case #%d: ",k);int s=1;while(s<tot&&ans[s]==0) s++;//注意不是小于等于tot,因为最终结果必须要输出一个数，比如90+10. for(int i=s;i<=len;i++) printf("%d",ans[i]);printf("\n");}return 0;}