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): 14    Accepted Submission(s): 8

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

2181000000000000

 

Sample Output

Case #1:299Case #2:29999999999991
Hint
9 + 9 = 18999999999999 + 1 = 1000000000000
 

 

Source

2016中国大学生程序设计竞赛（长春）-重现赛

 

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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 2004using namespace std;typedef long long LL;double anss;LL aans,sum;int cas,cass;int n,m,lll,ans;char s[N];int a[54][N],b[N],c[N];void gjdjian(int a[],int b[]){int i;for(i=1;i<=b[0];i++)a[i]-=b[i];for(i=1;i<=a[0];i++)if(a[i]<0)a[i]+=J,a[i+1]--;while(a[0]>1 && !a[a[0]])a[0]--;}void gjdprint(int a[]){int i;for(i=a[0];i;i--)printf("%d",a[i]);puts("");}void print(){int i,j;printf("Case #%d:\n",cass);printf("%d\n",lll);for(i=1;i<=lll;i++)gjdprint(a[i]);}int main(){#ifndef ONLINE_JUDGEW//freopen("1.txt","r",stdin);//freopen("2.txt","w",stdout);#endifint i,j,k;//init();//for(scanf("%d",&cass);cass;cass--)for(scanf("%d",&cas),cass=1;cass<=cas;cass++)//while(~scanf("%s",s))//while(~scanf("%d",&n)){lll=0;mem(a,0);scanf("%s",s);n=strlen(s);b[0]=n;for(i=0;i<n;i++)b[n-i]=s[i]-'0';while(!(b[0]==1 && b[1]==0)){if(b[0]==1){a[++lll][0]=1;a[lll][1]=b[1];break;}else if(b[0]==2 && b[2]==1){if(b[1]==0){a[++lll][0]=1;a[lll][1]=9;a[++lll][0]=1;a[lll][1]=1;break;}else if(b[1]==1){a[++lll][0]=2;a[lll][1]=1;a[lll][2]=1;break;}else{a[++lll][0]=2;a[lll][1]=1;a[lll][2]=1;a[++lll][0]=1;a[lll][1]=b[1]-1;break;}}else{for(i=b[0];i>b[0]/2;i--)c[i-b[0]/2]=b[i];c[0]=(b[0]+1)/2;int d[2]={1,1};gjdjian(c,d);j=c[0]+c[0];while(j>b[0])j--;lll++;a[lll][0]=j;for(i=c[0];i;i--,j--)a[lll][c[0]-i+1]=a[lll][j]=c[i];gjdjian(b,a[lll]);}}print();}return 0;}/*////*/