HDOJ 5920 Ugly Problem(构造+大数相减)

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Ugly Problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 812 Accepted Submission(s): 286
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 #2:29999999999991Hint
9 + 9 = 18999999999999 + 1 = 1000000000000

题意：将一个数s (1<=s<=10^1000) 拆成不超过50个回文数，这些回文数相加等于s。要求输出这样的数的个数，并从小到大输出这些数。

题解：每次都构造当前<=s的近似最大的回文数build ，然后 s=s-build 。用字符串表示的时候注意写法就行了。

代码如下：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define maxn 1010struct Node{char num[maxn];int len;}ans[102];//ans数组开在main函数厉害爆栈了 Node detail(Node str)//构造一个较大的回文数 {int temp=str.len/2+1;str.num[temp]--;for(int i=temp;i<=str.len;++i)if(str.num[i]<'0'){str.num[i+1]--;str.num[i]+=10;}if(str.num[str.len]=='0')str.len--;for(int i=1;i<=str.len/2;++i)str.num[i]=str.num[str.len-i+1];return str;}Node sub(Node res,Node subtor)//大数相减 {for(int i=1;i<=subtor.len;++i){if(res.num[i]>=subtor.num[i])res.num[i]=res.num[i]-subtor.num[i]+'0';else{res.num[i+1]--;res.num[i]+=10;res.num[i]=res.num[i]-subtor.num[i]+'0';}}for(int i=res.len;i>0;--i){if(res.num[i]=='0')res.len--;elsebreak;}return res;}int main(){int t,k=1;scanf("%d",&t);while(t--){Node str;memset(ans,0,sizeof(ans));scanf("%s",str.num+1);str.len=strlen(str.num+1);for(int i=1;i<=str.len/2;++i)//反转字符串，让高位表示高数位 swap(str.num[i],str.num[str.len-i+1]);int cnt=0;while(str.len>1){if(str.len==2&&((str.num[1]-'0'+(str.num[2]-'0')*10)<20)){ans[cnt].num[1]='9';ans[cnt++].len++;str=sub(str,ans[cnt-1]);continue;}Node build=detail(str);ans[cnt++]=build;str=sub(str,ans[cnt-1]);}ans[cnt++]=str;printf("Case #%d:\n%d\n",k++,cnt);for(int i=0;i<cnt;++i){for(int j=ans[i].len;j>=1;--j)putchar(ans[i].num[j]);printf("\n");}}return 0;}

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