【HDU5718 BestCoder 2nd AnniversaryA】【水题 高精度拆数相加】Oracle 正整数拆2正整数使得和最大

Oracle

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 536    Accepted Submission(s): 232


[align=left]Problem Description[/align]
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`.
 

[align=left]Input[/align]
The first line of the input contains an integer T (1≤T≤10),
which denotes the number of test cases.

For each test case, the single line contains an integer n (1≤n<1010000000).
 

[align=left]Output[/align]
For each test case, print a positive integer or a string `Uncertain`.
 

[align=left]Sample Input[/align]

3
112
233
1

 

[align=left]Sample Output[/align]

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.

 

[align=left]Source[/align]
BestCoder 2nd Anniversary
 

[align=left]Recommend[/align]
wange2014

#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,y,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
;
char b
;
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)

*/