Oracle【BC】

Oracle

Accepts: 631
Submissions: 2576

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

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 nn
without leading zeroes.

To get the meaning, he needs to rearrange the digits and split the number intotwo 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

.

Input

The first line of the input contains an integer T TT(1≤T≤10)
(1 \le T \le 10) (1≤T≤10),
which denotes the number of test cases.

For each test case, the single line contains an integer n nn(1≤n<1010000000)
(1 \le n < 10 ^ {10000000})(1≤n<10​10000000​​).

Output

For each test case, print a positive integer or a string

Uncertain

.

Sample Input
Copy

3
112
233
1

Sample Output
Copy

22
35
Uncertain

Hint
In the first example, it is optimal to split 112 112 112 into 21 21 21 and 1 1 1, and their sum is 21+1=22 21 + 1 = 22 21+1=22.

In the second example, it is optimal to split 233 233 233 into 2 2 2 and 33 33 33, and their sum is 2+33=35 2 + 33 = 35 2+33=35.

In the third example, it is impossible to split single digit 1 1 1 into two parts.

题意：把一串数分为两部分正数，要求和值最大，否则，输出”Uncertain”

分开判断 一位数、两位数，多位数则判断只有一位数不为0的情况，然后进行处理：sort升序排序，只要形成组合 sum = 最大的数+最小的正数【一位数不为0】

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
using namespace std;
const int maxn = 10000100;
int n, m, k;
char s[maxn],a[maxn];
int main(){
scanf("%d",&n);
while(n--){
memset(a,'\0',sizeof(a));
scanf("%s",s);
int len = strlen(s);
if(len==1){//一位数
printf("Uncertain\n");
continue;
}
if(len==2 ){//两位数
if(s[0]=='0'||s[1]=='0') printf("Uncertain\n");
else printf("%d\n",s[0]-'0'+s[1]-'0');
continue;
}
sort(s,s+len);//升序
k = 0;
int flag = 1;
int si = -1;
int cnt0 = 0;
for(int i=0;i<len;i++){
if(s[i]!='0' && flag){
si = i;//第一次不为0
flag = 0;
continue;
}
if(s[i]=='0')cnt0++;
a[k++] = s[i];
}
if(cnt0>=len-1) {//只有一个不为0的数
printf("Uncertain\n");
continue;
}
a[0] += s[si]-'0';
if(a[0] > '9'){
a[0] -= 10;
a[1] += 1;
}
m = k;
for(int i=1;i<k;i++){
if(i==k-1&&a[i]>'9') {
m++;
a[i] -= 10;
a[i+1] = '1';
continue;
}
if(a[i] > '9'){
a[i] -= 10;
a[i+1] += 1;
}
}
for(int i=m-1;i>=0;i--){
printf("%c",a[i]);
}
printf("\n");
}
return 0;
}