HDU 5920 Ugly Problem 【模拟】 (2016中国大学生程序设计竞赛(长春))

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

 

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题目链接：

　　http://acm.hdu.edu.cn/showproblem.php?pid=5920

题目大意：

　　输入一个长整数s（s<=10¹⁰⁰⁰），求将其拆分为不超过50个回文串之和的方案。

题目思路：

　　【模拟】

　　将前半段取出来，-1，构造成回文串c，s-=c，直到c=1。

　　特殊处理0~20的情况。

 //
 //by coolxxx
 //#include<bits/stdc++.h>
 #include<iostream>
 #include<algorithm>
 #include<string>
 #include<iomanip>
 #include<map>
 #include<stack>
 #include<queue>
 #include<set>
 #include<bitset>
 #include<memory.h>
 #include<time.h>
 #include<stdio.h>
 #include<stdlib.h>
 #include<string.h>
 //#include<stdbool.h>
 #include<math.h>
 #define min(a,b) ((a)<(b)?(a):(b))
 #define max(a,b) ((a)>(b)?(a):(b))
 #define abs(a) ((a)>0?(a):(-(a)))
 #define lowbit(a) (a&(-a))
 #define sqr(a) ((a)*(a))
 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
 #define mem(a,b) memset(a,b,sizeof(a))
 #define eps (1e-10)
 #define J 10
 #define mod 1000000007
 #define MAX 0x7f7f7f7f
 #define PI 3.14159265358979323
 #pragma comment(linker,"/STACK:1024000000,1024000000")
 #define N 2004
 using namespace std;
 typedef long long LL;
 double anss;
 LL aans,sum;
 int cas,cass;
 int n,m,lll,ans;
 char s[N];
 int a[][N],b[N],c[N];
 void gjdjian(int a[],int b[])
 {
     int i;
     for(i=;i<=b[];i++)
         a[i]-=b[i];
     for(i=;i<=a[];i++)
         if(a[i]<)
             a[i]+=J,a[i+]--;
     while(a[]> && !a[a[]])a[]--;
 }
 void gjdprint(int a[])
 {
     int i;
     for(i=a[];i;i--)
         printf("%d",a[i]);
     puts("");
 }
 void print()
 {
     int i,j;
     printf("Case #%d:\n",cass);
     printf("%d\n",lll);
     for(i=;i<=lll;i++)
         gjdprint(a[i]);
 }
 int main()
 {
     #ifndef ONLINE_JUDGEW
     freopen("1.txt","r",stdin);
 //    freopen("2.txt","w",stdout);
     #endif
     int i,j,k;
 //    init();
 //    for(scanf("%d",&cass);cass;cass--)
     for(scanf("%d",&cas),cass=;cass<=cas;cass++)
 //    while(~scanf("%s",s))
 //    while(~scanf("%d",&n))
     {
         lll=;mem(a,);
         scanf("%s",s);
         n=strlen(s);
         b[]=n;
         for(i=;i<n;i++)b[n-i]=s[i]-'';
         while(!(b[]== && b[]==))
         {
             if(b[]==)
             {
                 a[++lll][]=;
                 a[lll][]=b[];
                 break;
             }
             else if(b[]== && b[]==)
             {
                 if(b[]==)
                 {
                     a[++lll][]=;
                     a[lll][]=;
                     a[++lll][]=;
                     a[lll][]=;
                     break;
                 }
                 else if(b[]==)
                 {
                     a[++lll][]=;
                     a[lll][]=;
                     a[lll][]=;
                     break;
                 }
                 else
                 {
                     a[++lll][]=;
                     a[lll][]=;
                     a[lll][]=;
                     a[++lll][]=;
                     a[lll][]=b[]-;
                     break;
                 }
             }
             else
             {
                 for(i=b[];i>b[]/;i--)
                     c[i-b[]/]=b[i];
                 c[]=(b[]+)/;
                 int d[]={,};
                 gjdjian(c,d);
                 j=c[]+c[];
                 while(j>b[])j--;
                 lll++;
                 a[lll][]=j;
                 for(i=c[];i;i--,j--)
                     a[lll][c[]-i+]=a[lll][j]=c[i];
                 gjdjian(b,a[lll]);
             }
         }
         print();
     }
     return ;
 }
 /*
 //

 //
 */