HDU 5920 Ugly Problem（模拟）

这题思路比较简单，但是实现起来真的很烦……

比如一个数123456，先把它变成123321，剩下124，再变成121 和 3。

如果是654321，就先变成 653356 剩下的再递归。

特判一下10000000000这种情况，直接输出1和99999999999

这大概是我写过代码最长的模拟，撸了200多行……注意字符串手动剪切的话最后加个‘\0’；

【代码】

/* ***********************************************
Author :angon

************************************************ */
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <stack>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define showtime fprintf(stderr,"time = %.15f\n",clock() / (double)CLOCKS_PER_SEC)
#define lld %I64d
#define REP(i,k,n) for(int i=k;i<n;i++)
#define REPP(i,k,n) for(int i=k;i<=n;i++)
#define scan(d) scanf("%d",&d)
#define scanl(d) scanf("%I64d",&d)
#define scann(n,m) scanf("%d%d",&n,&m)
#define scannl(n,m) scanf("%I64d%I64d",&n,&m)
#define mst(a,k) memset(a,k,sizeof(a))
#define LL long long
#define N 1500
#define mod 1000000007
inline int read(){int s=0;char ch=getchar();for(; ch<'0'||ch>'9'; ch=getchar());for(; ch>='0'&&ch<='9'; ch=getchar())s=s*10+ch-'0';return s;}

struct Bignum
{
char s
;
};
int a
,b
,c
;
Bignum operator - (Bignum q, Bignum p)
{
Bignum ans;
int la = strlen(q.s);
int lb = strlen(p.s);
mst(c,0);mst(a,0);mst(b,0);
for(int i = 0;i<la;i++)
a[i] = q.s[i] - '0';
for(int i = 0;i<lb;i++)
b[i] = p.s[i] - '0';
for(int i = la-1,j=lb-1;i>=0;i--,j--)
{
if(j<0)
{
c[i] = a[i];
}
if(a[i] >= b[j])
{
c[i] = a[i] - b[j];
}
else
{
a[i] += 10;
a[i-1] -= 1;
c[i] = a[i] - b[j];
}
}
int i = 0;
while(c[i]==0) i++;
mst(ans.s,0);
int j;
for(j=0;i<la;i++,j++)
{
ans.s[j] = c[i] + '0';
}
ans.s[j] = '\0';
return ans;
}

bool operator >= (Bignum q,Bignum p)
{
int la = strlen(q.s);
int lb = strlen(p.s);
if(la>lb) return 1;
if(la<lb) return 0;
for(int i=0;i<la;i++)
if(q.s[i]==p.s[i]){}
else if(q.s[i]>p.s[i]) return 1;
else if(q.s[i]<p.s[i]) return 0;
return 1;
}
string ans[55];
Bignum get_l(Bignum n)
{
Bignum q; mst(q.s,0);
int len = strlen(n.s);
int j,i;
for(i=0,j=0;i<len/2;i++,j++)
{
q.s[j] = n.s[i];
}
q.s[j] = '\0';
return q;
}
Bignum get_r(Bignum n)
{
Bignum q; mst(q.s,0);
int len = strlen(n.s);
int i,j;
for(i=(len+1)/2,j=0;i<len;i++,j++)
{
q.s[j] = n.s[i];
}
q.s[j] = '\0';
return q;
}

int num;
void dfs(Bignum x)
{
int len = strlen(x.s);
bool flag = 0;
for(int i =0;i<len;i++)
{
if(x.s[i] != x.s[len - i -1])
{
flag = 1;
break;
}
}
if(!flag)
{
ans[num++] = (string)x.s + '\0';
return ;
}
flag = 0;
if(x.s[0] == '1')
{
for(int i=1;i<len;i++)
if(x.s[i] != '0')
{
flag = 1;
break;
}
if(!flag)
{
Bignum tmp ;
strcpy(tmp.s ,"1");
ans[num++] = "1";
ans[num++] = (string)(x - tmp).s + '\0';;
return ;
}
}
if(len == 1)
{
if(x.s[0]!='0')
ans[num++] = x.s;
return ;
}
Bignum l = get_l(x);
Bignum r = get_r(x);
Bignum newl ;
for(int i=len/2-1;i>=0;i--)
{
newl.s[len/2-1 - i] = l.s[i];
}
if(r>=newl)
{
if(len & 1)
{
ans[num++] = (string)l.s + x.s[len/2] + (string)newl.s + '\0';;
}
else
{
ans[num++] = (string)l.s + (string)newl.s + '\0';;
}
//cout<< (r - newl).s<<endl;
dfs(r - newl);
}
else
{
Bignum tmp1,tmp;
strcpy(tmp1.s ,"1");
Bignum res = l - tmp1;
mst(newl.s,0);
for(int i=len/2-1;i>=0;i--)
{
newl.s[len/2-1 - i] = res.s[i];
}
if(len & 1)
{
ans[num++] = (string)res.s + x.s[len/2] + (string)newl.s + '\0';;
for(int i=0;i<len;i++)
tmp.s[i] = ans[num-1][i];
tmp.s[len] = '\0';
}
else
{
ans[num++] = (string)res.s + (string)newl.s + '\0';;
for(int i=0;i<len;i++)
tmp.s[i] = ans[num-1][i];
tmp.s[len] = '\0';
}
Bignum tt = x - tmp;
dfs(tt);
}
}

int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
Bignum n;
int t,cas=1;
scan(t);
while(t--)
{
mst(n.s,0);
cin>>n.s;
REP(i,0,55) ans[i].clear();
num = 0;
dfs(n);
printf("Case #%d:\n",cas++);
printf("%d\n",num);
REP(i,0,num)
cout<<ans[i]<<'\n';
}
return 0;
}