D - Ugly Problem HDU - 5920

D - Ugly Problem

HDU - 5920

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.

InputIn 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).OutputFor 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

OJ-ID：
hdu-5920

author：
Caution_X

date of submission：
20191029

tags：
模拟

description modelling：
给定一个数n，把n拆成若干个数相加，且这若干个数是回文串
输出：输出拆成了多少个数以及每一个数的大小

major steps to solve it：
把一个数随机拆成两个回文串显然十分不现实。
我们可以把一个按照它所能拆的最大回文串来拆：n=n1(回文串)+n2
若n2不是回文串，n2代替n的位置继续拆出新数，若n2也是回文串，结束，输出答案。

warnings：
n=10特判

AC code:

#include<bits/stdc++.h>
using namespace std;
char num[],sub[];
char ans[][];
char one[]="";
bool judge(char *s){//判断当前数字是否为回文数字
   int lens = strlen(s);
   for(int i = ;i<lens/;i++){
        if(s[i]!=s[lens - i - ]){
            return false;
        }
   }
   return true;
}
void decrease(char *s1,char *s2)
{
    int len1=strlen(s1);
    int len2=strlen(s2);
    int i=len1-,j=len2-,flag=;
    while(i>=&&j>=) {
        if(s1[i]-''-flag>=) {
            s1[i]-=flag;
            flag=;
        }
        else {
            s1[i] = s1[i] +  - flag;
            flag = ;
        }
        if(s1[i]>=s2[j]) {
            s1[i]=s1[i]-s2[j]+'';
        }
        else {
            s1[i]=s1[i]+-s2[j]+'';
            flag=;
        }
        i--,j--;
    }
    while(i>=&&flag) {
        if(s1[i]-''>=flag) {
            s1[i]-=flag;
            flag=;
        }
        else {
            s1[i]=s1[i]+-flag;
            flag=;
        }
        i--;
    }
    int id=;
    bool pre_0=true;
    char tmp[];
    memset(tmp,,sizeof(tmp));
    for(int k=;k<len1;k++) {
        if(s1[k]==''&&pre_0)    continue;
        if(s1[k]!=''&&pre_0) pre_0=false;
        tmp[id++]=s1[k];
    }
    if(!id) {
        tmp[id++]='';
    }
    tmp[id]='\0';
    strcpy(s1,tmp);
}
void palindromic(char *s1,char *s2)
{
    int len1=strlen(s1),len2;
    if(len1==&&s1[]==''&&s1[]==''){
        s2[]='';
        s2[]='\0';
        return;
    }
    if(len1&) len2=len1/+;
    else len2=len1/;
    for(int i=;i<len2;i++) {
        s2[i]=s1[i];
    }
    s2[len2]='\0';
    decrease(s2,one);
    if(s2[]=='') {
        s2[]='';
    }
    for(int i=len1-,j=;j<i;j++,i--) {
        s2[i]=s2[j];
    }
    s2[len1]='\0';
}
int main()
{
    //freopen("input.txt","r",stdin);
    int T,kase=;
    scanf("%d",&T);
    while(T--) {
        scanf("%s",num);
        int id=;
        while(num[]!=''&&id<) {
            if(judge(num)) {
                strcpy(ans[id++],num);
                break;
            }
            memset(sub,,sizeof(sub));
            palindromic(num,sub);
            strcpy(ans[id++],sub);
            decrease(num,sub);
        }
        printf("Case #%d:\n%d\n",kase++,id);
        for(int i=;i<id;i++) {
            printf("%s\n",ans[i]);
        }
    }
}