D - Ugly Problem
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.
InputIn 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).OutputFor 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
OJ-ID：
hdu-5920
author：
Caution_X
date of submission：
20191029
tags：
模拟
description modelling：
给定一个数n，把n拆成若干个数相加，且这若干个数是回文串
输出：输出拆成了多少个数以及每一个数的大小
major steps to solve it：
把一个数随机拆成两个回文串显然十分不现实。
我们可以把一个按照它所能拆的最大回文串来拆：n=n1(回文串)+n2
若n2不是回文串，n2代替n的位置继续拆出新数，若n2也是回文串，结束，输出答案。
warnings：
n=10特判
AC code:
#include<bits/stdc++.h>
using namespace std;
char num[1005],sub[1005];
char ans[55][1005];
char one[2]="1";
bool judge(char *s){//判断当前数字是否为回文数字
int lens = strlen(s);
for(int i = 0;i<lens/2;i++){
if(s[i]!=s[lens - i - 1]){
return false;
}
}
return true;
}
void decrease(char *s1,char *s2)
{
int len1=strlen(s1);
int len2=strlen(s2);
int i=len1-1,j=len2-1,flag=0;
while(i>=0&&j>=0) {
if(s1[i]-‘0‘-flag>=0) {
s1[i]-=flag;
flag=0;
}
else {
s1[i] = s1[i] + 10 - flag;
flag = 1;
}
if(s1[i]>=s2[j]) {
s1[i]=s1[i]-s2[j]+‘0‘;
}
else {
s1[i]=s1[i]+10-s2[j]+‘0‘;
flag=1;
}
i--,j--;
}
while(i>=0&&flag) {
if(s1[i]-‘0‘>=flag) {
s1[i]-=flag;
flag=0;
}
else {
s1[i]=s1[i]+10-flag;
flag=1;
}
i--;
}
int id=0;
bool pre_0=true;
char tmp[1005];
memset(tmp,0,sizeof(tmp));
for(int k=0;k<len1;k++) {
if(s1[k]==‘0‘&&pre_0) continue;
if(s1[k]!=‘0‘&&pre_0) pre_0=false;
tmp[id++]=s1[k];
}
if(!id) {
tmp[id++]=‘0‘;
}
tmp[id]=‘\0‘;
strcpy(s1,tmp);
}
void palindromic(char *s1,char *s2)
{
int len1=strlen(s1),len2;
if(len1==2&&s1[0]==‘1‘&&s1[1]==‘0‘){
s2[0]=‘9‘;
s2[1]=‘\0‘;
return;
}
if(len1&1) len2=len1/2+1;
else len2=len1/2;
for(int i=0;i<len2;i++) {
s2[i]=s1[i];
}
s2[len2]=‘\0‘;
decrease(s2,one);
if(s2[0]==‘0‘) {
s2[0]=‘1‘;
}
for(int i=len1-1,j=0;j<i;j++,i--) {
s2[i]=s2[j];
}
s2[len1]=‘\0‘;
}
int main()
{
//freopen("input.txt","r",stdin);
int T,kase=1;
scanf("%d",&T);
while(T--) {
scanf("%s",num);
int id=0;
while(num[0]!=‘0‘&&id<50) {
if(judge(num)) {
strcpy(ans[id++],num);
break;
}
memset(sub,0,sizeof(sub));
palindromic(num,sub);
strcpy(ans[id++],sub);
decrease(num,sub);
}
printf("Case #%d:\n%d\n",kase++,id);
for(int i=0;i<id;i++) {
printf("%s\n",ans[i]);
}
}
}
原文地址：https://www.cnblogs.com/cautx/p/11762119.html