HDU 5718 Oracle（大数）

Oracle

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1235    Accepted Submission(s): 535

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

31122331

 

Sample Output

2235Uncertain
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：

给你一串数，让你任意组合这些数，变成两个【正数】，求相加和最大。

不能变成则输出“”

POINT：

只要这串数有2个以上非0的数，就可以拆。

拆肯定是拿一个最小的非0的数。加上一个9998877……00排列的数。

模拟大数加一下就好

#include <stdio.h>#include <map>#include<iostream>#include <math.h>#include <algorithm>#include <string.h>#include <vector>#include <queue>#include <set>using namespace std;typedef long long  LL;const int maxn = 10001000;char s[maxn];int num[20];int a[maxn];int main(){    int T;    scanf("%d",&T);    while(T--){        memset(num,0,sizeof num);        scanf("%s",s);        int l=strlen(s);        for(int i=0;i<l;i++){            num[s[i]-'0']++;        }        if(num[0]>=l-1){            printf("Uncertain\n");            continue;        }        int k;        for(int i=1;i<=9;i++){            if(num[i]>=1){                k=i;                num[i]--;                break;            }        }        for(int i=0;i<=l;i++) a[i]=0;        int now=0;        for(int i=0;i<=9;i++){            for(int j=1;j<=num[i];j++){                a[++now]=i;            }        }        a[1]+=k;        for(int i=1;i<=now;i++){            if(a[i]>=10){                a[i+1]+=a[i]/10;                a[i]%=10;                if(i==now){                    now++;                    break;                }            }else{                break;            }        }        for(int i=now;i>=1;i--){            printf("%d",a[i]);        }        printf("\n");    }    return 0;}