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题意：一串电话号码，每个数字+8取各位后，把每个数字写成对应的大写英文，从"ZERO"和“NINE”，然后打乱字母的顺序。现在给出打乱的字母顺序，问可能的字典序最小的电话号码是是多少（可能有前导0）
思路：分析0..9 每个数字的英文组成。。。然后大概类似解方程。。可以根据字母的个数确定每个数字的个数。。。
然后-8。。。存一下排个序就好了。。。1A
代码实现
1#include <cstdio>
2#include <cstring>
3#include <iostream>
4#include <algorithm>
5#include <vector>
6#include <queue>
7#include <set>
8#include <map>
9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair
21using namespace std;
22const double eps = 1E-8;
23const int dx4[4]={1,0,0,-1};
24const int dy4[4]={0,-1,1,0};
25const int inf = 0x3f3f3f3f;
26const int N=1E4+7;
27char st[N];
28int len;
29int a[30];
30int cnt[15]; //0..9数字的个数.
31int main()
32{
33 int T;
34 cin>>T;
35 while (T--)
36 {
37 scanf("%s",st);
38 len = strlen(st);
39 ms(a,0);
40 ms(cnt,0);
41 for ( int i = 0 ; i < len ; i++)
42 {
43 int val = st[i]-'A'+1;
44 a[val]++;
45 }
46 cnt[0] = a['Z'-'A'+1];
47 cnt[6] = a['X'-'A'+1];
48 cnt[2] = a['W'-'A'+1];
49 cnt[8] = a['G'-'A'+1];
50 cnt[4] = a['U'-'A'+1];
51 cnt[7] = a['S'-'A'+1]-cnt[6];
52 cnt[5] = a['V'-'A'+1]-cnt[7];
53 cnt[1] = a['O'-'A'+1]-cnt[0]-cnt[2]-cnt[4];
54 cnt[3] = a['H'-'A'+1]-cnt[8];
55 cnt[9] = (a['N'-'A'+1]-cnt[1]-cnt[7])/2;
56 vector <int>ans;
57 for ( int i = 0 ; i <= 9 ; i++)
58 {
59 for ( int j = 1 ; j <= cnt[i] ; j++)
60 {
61 int val = i-8;
62 if (val<0) val+=10;
63 ans.push_back(val);
64 }
65 }
66 sort(ans.begin(),ans.end());
67 int siz = ans.size();
68 for ( int i = 0 ; i < siz; i ++) printf("%d",ans[i]);
69 }
70 #ifndef ONLINE_JUDGE
71 fclose(stdin);
72 #endif
73 return 0;
74}