ccpc 长春站 J 题 Ugly Problem

Ugly Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 0    Accepted Submission(s): 0
Special Judge

Problem Description

Everyone hates ugly problems.

You are given a positive integer. You must represent that number by sum of palindromic numbers.

A palindromic number is a positive integer such that if you write out that integer as a string in decimal without leading zeros, the string is an palindrome. For example, 1 is a palindromic number and 10 is not.

 

Input

In the first line of input, there is an integer T denoting the number of test cases.

For each test case, there is only one line describing the given integer s (1≤s≤101000).

 

Output

For each test case, output “Case #x:” on the first line where x is the number of that test case starting from 1. Then output the number of palindromic numbers you used, n, on one line. n must be no more than 50. en output n lines, each containing one of your
palindromic numbers. Their sum must be exactly s.

 

Sample Input

2
18
1000000000000

 

Sample Output

Case #1:
2
9
9
Case #2:
2
999999999999
1

Hint
9 + 9 = 18
999999999999 + 1 = 1000000000000
题意很简单，就是给一个数，输出这个数可以由多少个回文串加和得到，输出个数以及每个回文串，回文串个数不得高于50 解：我们取当前串的前面一半，例如 93840 ，取938 ，然后另这串减1 ，（减1是为了对称后比原来串小）得937，然后再对称得到 93739 ，用原串减去当前串得到一个位数减少一半的新串 111，然后再采取这样的操作，直到当前串变为一位时，退出循环，每一次的对称得到的回文串就是一个答案，记录回文串个数及回文串输出即可
特殊判断 当前串的奇偶，奇数取前面一半加中间位，偶数位取前面一半位；当取得的数为两位数且十位为1时也要特殊判断，因为取得的前面一半串为1，减1 以后的0，对称以后为00 ，一个串减去00 永远不变，会造成无限循环，所以这时我们特殊处理一下，当当前串十位比个位大时取得的前面一半数不减1，直接对称过来， 比如 当前串为15 那么 取 1 对称得11，做差得4 ，变为1位数退出。当当前串十位小于等于个位就两种情况10 11 ，那么不再对称，直接减去9，剩余1 2 ，变为1位数退出循环。

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn=1e6+10;

bool comp(string num1,string num2)
{
int leng=num1.length(),i;
for(i=0; i<leng; i++)
{
if(num1[i]!='0')break;
}
num1=num1.substr(i,leng);
if(num1.length()==0)num1="0";

leng=num2.length();
for(i=0; i<leng; i++)
{
if(num2[i]!='0')break;
}
num2=num2.substr(i,leng);
if(num2.length()==0)num2="0";

if(num1.length()>num2.length())return true;
else if(num1.length()==num2.length())
{
if(num1>=num2)return true;
else return false;
}
else
return false;
}
string _minus(string num1,string num2)
{
if(comp(num2,num1))
{
string ss=num1;
num1=num2;
num2=ss;
}
reverse(num1.begin(),num1.end());
reverse(num2.begin(),num2.end());
string result="";

int i;
for(i=0; i<num1.length()&&i<num2.length(); i++)
{
char c=num1[i]-num2[i]+48;
result=result+c;
}
if(i<num1.length())for(; i<num1.length(); i++)result=result+num1[i];

int jiewei=0;
for(i=0; i<result.length(); i++)
{
int zhi=result[i]-48+jiewei;
if(zhi<0)
{
zhi=zhi+10;
jiewei=-1;
}
else jiewei=0;
result[i]=(char)(zhi+48);
}

for(i=result.length()-1; i>=0; i--)
{
if(result[i]!='0')break;
}

result=result.substr(0,i+1);
reverse(result.begin(),result.end());
if(result.length()==0)result="0";
return result;
}
vector<string>ans;
string s1;
string fen(string str)
{
s1="";
string s2;
int len=str.length();
if(len&1)
{
for(int i=0; i<=len/2; i++)
{
s1+=str[i];
}
string temp="1";
s1=_minus(s1,temp);
for(int j=len/2-1, i=len/2+1; i<len; i++,j--)
{
s1+=s1[j];
}
s2=_minus(str,s1);
}
else
{
if(str=="11"||str=="10")/// 十位数字小于等于个位
{
s1="9";
return _minus(str,s1);
}
for(int i=0; i<len/2; i++)
s1+=str[i];
if(!(len==2&&str[0]=='1'))
{
string temp="1";
s1=_minus(s1,temp);
}
///两位数字，十位数字为1，且十位数字小于个位，不再减一，直接对称过去
for(int j=len/2-1, i=len/2; i<len; i++,j--)
s1+=s1[j];

s2=_minus(str,s1);
}
return s2;
}
bool cmp(string a,string b)
{
return a>b;
}
int main()
{
int T;
string str,ss,strs;
scanf("%d",&T);
for(int cas=1; cas<=T; cas++)
{
ans.clear();
cin>>str;
printf("Case #%d:\n",cas);
int tot=0;
while(1)
{
string temp="";
if(str.length()==1)
{
ans.push_back(str);
str="";
break;
}
temp+=fen(str);
str="";
str+=temp;
if(str.length()==0)
break;
ans.push_back(s1);
}
tot=ans.size();
cout<<tot<<endl;
for(int i=0; i<tot; i++)
{
cout<<ans[i]<<endl;
}
}
return 0;
}