BestCoder 2nd Anniversary Oracle

最新推荐文章于 2016-12-08 21:19:13 发布

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本文介绍了一个关于如何从一个整数通过重新排列其数字并分成两个正整数来获得最大可能总和的问题。文章详细解释了输入输出的要求，并提供了一个具体的示例代码实现。

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Oracle

Accepts: 599

Submissions: 2576

Time Limit: 8000/4000 MS (Java/Others)

Memory Limit: 262144/262144 K (Java/Others)

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 two positive integers without leading zeroes, 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.

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<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 10000010;
char s[maxn];
int b[maxn];
int a[maxn];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",s);
        int len = strlen(s);
        int p = 0;
        for(int i = 0 ; i < len ; i++)
        {
            a[i] = s[i] - '0';
            if(a[i] != 0)
                p++;
        }
        if(p <= 1)
        {
            printf("Uncertain\n");
            continue;
        }
        sort(a,a+len);
        int k;
        for(int i = 0 ; i < len ; i++)
        {
            if(a[i] != 0)
            {
                k = i;
                break;
            }
        }
        int l = 0;
        for(int i = 0 ; i < len ; i++)
        {
            if(i == k) continue;
            b[l] = a[i];
            l++;
        }
        b[0] += a[k];
        b[l] = 0;
        for(int j = 0 ; j < l ; j++)
        {
            if(b[j] < 10) break;
            b[j] %= 10;
            b[j+1] += b[j] / 10;
        }
        if(b[l] != 0)
            l++;
        for(int i = l-1 ; i >= 0 ; i--)
            printf("%d",b[i]);
        printf("\n");
    }
    return 0;
}