HDU 5718 Oracle

Oracle

Time Limit : 8000/4000ms (Java/Other)   Memory Limit : 262144/262144K (Java/Other)

Total Submission(s) : 11   Accepted Submission(s) : 4

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Problem 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≤T≤10), which denotes the number of test cases.

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

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.

没有前置零，因此像1000这样的数字也是不能拆的，还有两位数需要特殊考虑。其他的情况只需找出最小的非零数，

其他的数字按降序排列形成一个数字，两数相加即可

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string.h>
int n[10], v[10000002],q,L;
char a[10000002];
using namespace std;
void pp(int p,int k)
{
    if (k == L - 1)
    {
        L++;
        if (p + v[k] <= 9) v[k] = v[k] + p;
        else
        {
            v[k] = 0; v[k+1] = 1;
        }
        return;
    }
    if (p + v[k] <= 9) v[k] = v[k] + p;
    else
    {
        int y = v[k] + p; v[k] = y % 10;
        pp(1, k + 1);
    }
}
int main()
{
    int T; scanf("%d", &T);
    while (T--)
    {
        scanf("%s", &a);
        memset(n, 0, sizeof(n));
        L = strlen(a);
        for (int i = 0; i < L; i++)
        {
            n[a[i] - '0']++;
        }
        if (L == 1 || n[0] == L - 1)
        {
            printf("Uncertain\n");
            continue;
        }
        for (int i = 1; i <= 9; i++)
        {
            if (n[i])
            {
                q = i, n[i]--;
                break;
            }
        }
        if (L == 2)
        {
            for (int i = 0; i <= 9; i++)
            {
                if (n[i])
                {
                    printf("%d\n", q + i);
                    break;
                }
            }
            continue;
        }
        int t = 1;
        for (int i = 0; i <= 9; i++)
        {
            for (int j = 1; j <= n[i]; j++)
            {
                v[t] = i; t++;
            }
        }
        pp(q,1);
        for (int i = L - 1; i >= 1; i--)
            printf("%d", v[i]);
        printf("\n");
    }
    return 0;
}