HDU - 5718 - Oracle

Oracle

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1232    Accepted Submission(s): 532

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.
 
   
    
  

 

Source

BestCoder 2nd Anniversary 

 

题意：有一长度为n（ n ≤ 1e7 ）的无前置零数字，求重新布置该数字的每一位数之后，将其分为两部分的和最大为多少，分出的数字不能有前置零

思路：先排序，要使得和最大，其中一个数要为个位数（尽量小且不为0），剩下的都是另一个数

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define LL long long
const int N = 1e7 + 10;
LL T,n,len,l;
char s[N],ans[N],zero[N];
bool cmp(int a,int b) {return a>b;}
int main()
{
    scanf("%lld",&T);
    for(int i=0;i<N;i++) zero[i] = '0';
    while(T--){
        scanf("%s",s+1);
        len = strlen(s+1);
        sort(s+1,s+len+1,cmp);//从大到小排序
        if(len==1||s[2]=='0') {
            //当整个数字中的非零数少于2个时不可分为两个部分
            printf("Uncertain\n");
            continue;
        }
        //最小数为0时
        if(s[len]=='0'){
            LL pos = -1;
            //找出最小非零数所在位置
            for(LL i=len;i>=1;i--){
                if(s[i]!='0'){
                    pos = i;break;
                }
            }
            char c = s[pos];
            s[pos] = s[len] = 0;
            printf("%s%s%c\n",s+1,s+pos+1,c);
        }else{
            LL pos = -1;
            //找位数最低的非9数 --- 进位只会持续到该位置，结束后原来的9变为0
            for(LL i=len-2;i>=1;i--){
                if(s[i]!='9'){
                    pos = i;break;
                }
            }
            int num1 = s[len] - '0',num2 = s[len-1] -'0';
            //两数末尾数字
            if(num1+num2>=10){
                //产生进位
                if(pos==-1){
                    //全为9
                    zero[len-2] = 0;
                    printf("1%s%d\n",zero,(num1+num2)%10);
                    zero[len-2] = '0';
                }else{
                    zero[len-pos-2] = 0;
                    int n = s[pos] - '0' + 1;
                    s[pos] = 0;
                    printf("%s%d%s%d\n",s+1,n,zero,(num1+num2)%10);
                    zero[len-pos-2] = '0';
                }
            }else{
                //不产生进位
                s[len-1] = 0;
                printf("%s%d\n",s+1,num1+num2);
            }
        }
    }
    return 0;
}