Oracle

  
  
   
   3
112
233
1
  
  

  
  
   
   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. 
   
   
    
  
  

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in","r",stdin); freopen("c://test//output.out","w",stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1,class T2>inline void gmax(T1 &a,T2 b) { if (b>a)a = b; }
template <class T1,class T2>inline void gmin(T1 &a,T2 b) { if (b<a)a = b; }
const int N = 1e7+10,M = 0,Z = 1e9 + 7,ms63 = 0x3f3f3f3f;
int casenum,casei;
char s[N];
char b[N];
int cnt[10];
int main()
{
	scanf("%d",&casenum);
	for (casei = 1; casei <= casenum; ++casei)
	{
		scanf("%s",s); int n = strlen(s);
		MS(cnt,0);
		for (int i = 0; s[i]; ++i)++cnt[s[i] - 48];
		int no0 = 0;
		for (int i = 1; i <= 9; ++i)no0 += cnt[i];
		if (no0 < 2)puts("Uncertain");
		else
		{
			int plus = 0;
			for (int i = 1; i <= 9; ++i)if (cnt[i])
			{
				--cnt[i];
				plus = i;
				break;
			}
			int m = 0;
			for (int i = 0; i <= 9; ++i)
			{
				while(cnt[i]--)b[m++] = 48 + i;
			}
			b[m] = 0;
			b[0] += plus;
			int p = 0;
			while (b[p] > '9')
			{
				b[p] -= 10;
				++b[p + 1];
				if (b[p + 1] == 1)b[p + 1] += 48;
				++p;
			}
			if (p == m)b[m++] = 0;
			reverse(b,b + m);
			puts(b);
		}
	}
	return 0;
}
/*
【trick&&吐槽】
 用char作为输入输出会快很多

【题意】
给你一个正整数，长度在1e7范围内。
让你把这个正整数拆成2个正整数（无法做到输出Uncertain)，使其和尽可能大

【类型】
水题

【分析】
显然，我们使得A尽可能大，使得B为最小正整数，然后求和即可。

【时间复杂度&&优化】
O(Tn)

*/

（编辑：李大同）

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