HDOJ 5920 Ugly Problem(构造+大数相减)
发布时间：2021-01-21 08:04:11 所属栏目：大数据 来源：网络整理
导读：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 positi
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Ugly ProblemTime 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 ( 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
?
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 1010
struct 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--;
else break;
}
return res;
}
int main()
{
int t,k=1;
scanf("%d",&t);
while(t--)
{
Node str;
memset(ans,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%dn",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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