Time Limit:4000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Description
There is once a king and queen, rulers of an unnamed city, who have three daughters of conspicuous beauty.
The youngest and most beautiful is Psyche, whose admirers, neglecting the proper worship of the love goddess Venus, instead pray and make offerings to her. Her father, the king, is desperate to know about her destiny, so he comes to the Delphi Temple to ask for an oracle.
The oracle is an integer $ n $ without leading zeroes.
To get the meaning, he needs to rearrange the digits and split the number into <b>two positive integers without leading zeroes</b>, and their sum should be as large as possible.
Help him to work out the maximum sum. It might be impossible to do that. If so, print `Uncertain`.
Input
The first line of the input contains an integer $ T $ $ (1 \le T \le 10) $, which denotes the number of test cases.
For each test case, the single line contains an integer $ n $ $ (1 \le n < 10 ^ {10000000}) $.
Output
For each test case, print a positive integer or a string `Uncertain`.
Sample Input
3 112 233 1
Sample Output
22 35 Uncertain
Hint
In the first example, it is optimal to split $ 112 $ into $ 21 $ and $ 1 $, and their sum is $ 21 + 1 = 22 $. In the second example, it is optimal to split $ 233 $ into $ 2 $ and $ 33 $, and their sum is $ 2 + 33 = 35 $. In the third example, it is impossible to split single digit $ 1 $ into two parts.
从给出的数中挑一个非零最小数出来，主要是高精度的处理
1 #include<cstdio>
2 #include<iostream>
3 #include<cstring>
4 #include<stdlib.h>
5 #include<algorithm>
6 using namespace std;
7 #define N 10000005
8 char s[N];
9 int main()
10 {
11 int T,len,minn,mi,i;
12 scanf("%d",&T);
13 while(T--)
14 {
15 scanf("%s",&s);
16 len=strlen(s);
17 sort(s,s+len);
18 if(len==1||s[len-2]==‘0‘)
19 printf("Uncertain\n");
20 else
21 {
22 for( i=0;i<len;i++)
23 if(s[i]!=‘0‘)
24 {
25 minn=i;
26 mi=s[i]-‘0‘;
27 break;
28 }
29 for(i=minn;i>0;i--)
30 s[i]=s[i-1];//接下来都是从1开始
31 s[1]+=mi;
32 for(i=1;i<len-1;i++)
33 {
34 if(s[i]>‘9‘)
35 {
36 s[i+1]++;
37 s[i]-=10;
38 }
39 else
40 break;
41 }
42 printf("%d",s[len-1]-‘0‘);
43 for(i=len-2;i>=1;i--)
44 printf("%c",s[i]);
45 printf("\n");
46 }
47 }
48 return 0;
49 }