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题意:将数字n转化成若干个回文数字之和。
知识点:大数减法 前缀0删除 字符串指针的应有
题解:因为带着完美主义的追求，把这题做傻了。。。
思路是对的，只是多考虑了些不必要的细节，导致本来简单的问题复杂化了,又被指针坑了一把。
所幸AC了，就贴出来当作对指针的回顾吧。
简洁解法: 将s的前半段s2取出来，然后s2-1，再构造成回文串c。s-=c，直到c=1。这么做简单不少，看来比赛时不能太追求完美，AC为王。
#include"bits/stdc++.h"
using namespace std;
const int N=1e3+100;
char *s,*small; //指向字符串的指针(为了快速删除前缀0而如此定义)
char ss[N],sm[N]; //给字符串指针 分配初始空间 void delete0(char * &ps) //删除前缀0
{for(int i=0;i<strlen(ps);i++){if(ps[i]!='0'){ps=&ps[i];return ;}}ps[0]=0;
}void sub(int lens) //大数减法
{delete0(s);delete0(small);bool flag=0;int cha=strlen(s)-strlen(small);for(int i=lens-1;i>=0;i--){if(flag==1){if(s[i]=='0')s[i]='9';elses[i]=s[i]-1,flag=0;}if(i-cha<0)break;if(s[i]>=small[i-cha])s[i]=s[i]-small[i-cha]+'0';else{s[i]=s[i]+10-small[i-cha]+'0';flag=1;}}
}void find_s()
{int lens=strlen(s);if(lens%2==0){if(s[lens/2-1]>=s[lens/2]) //中间左大于右 其实这不考率也行，毕竟n>50 {int key;for(int i=lens/2-1;i>=0;i--)if(s[i]!='0') // 0特殊考虑{key=i;break;}if(key==0){small[0]='0';for(int i=1;i<lens;i++){small[i]='9';}return ;}for(int i=0;i<=key;i++){small[i]=small[lens-i-1]=s[i];if(i==key)small[i]=small[lens-i-1]=s[i]-1;}for(int i=key+1;i<lens/2;i++)small[i]=small[lens-i-1]='9';}else{for(int i=0;i<lens/2;i++)small[i]=small[lens-i-1]=s[i];}}else //lens奇数{if(lens==1){small[0]=s[0];return ;}if(s[(lens+1)/2-1]!='0'){for(int i=0;i<(lens+1)/2;i++){small[i]=small[lens-i-1]=s[i];if(i==(lens+1)/2-1)small[i]=s[i]-1;}}else ///和偶数时的处理一样{if(s[(lens+1)/2-2]>=s[(lens+1)/2]) //中间左大于右 {int key;for(int i=(lens+1)/2-2;i>=0;i--)if(s[i]!='0'){key=i;break;}if(key==0){small[0]='0';for(int i=1;i<lens;i++){small[i]='9';}return ;}for(int i=0;i<=key;i++){small[i]=small[lens-i-1]=s[i];if(i==key)small[i]=small[lens-i-1]=s[i]-1;}for(int i=key+1;i<(lens+1)/2;i++)small[i]=small[lens-i-1]='9'; }else{for(int i=0;i<lens/2;i++)small[i]=small[lens-i-1]=s[i];small[lens/2]='0';}}}
}char ans[60][N];
int main()
{int t;scanf("%d",&t);for(int kase=1;kase<=t;kase++){memset(ans,0,sizeof(ans));scanf("%s",ss);s=ss;small=sm;int lens,cnt=0;while(true){memset(sm,0,sizeof(sm));small=sm; //small指针的初始化delete0(s);lens=strlen(s);if(s[0]=='0'||lens==0)break;find_s();delete0(small);for(int i=0;i<strlen(small);i++)ans[cnt][i]=small[i];cnt++;sub(lens);}printf("Case #%d:\n",kase);printf("%d\n",cnt);for(int i=0;i<cnt;i++)printf("%s\n",ans[i]);}
}Ugly Problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 385 Accepted Submission(s): 145
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.
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).
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 1Hint9 + 9 = 18 999999999999 + 1 = 1000000000000
Source
2016中国大学生程序设计竞赛（长春）-重现赛