hdu 5651 xiaoxin juju needs help

Posted 2020-07-03 lucky_少哖

xiaoxin juju needs help

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1505    Accepted Submission(s): 437

Problem Description

As we all known, xiaoxin is a brilliant coder. He knew **palindromic** strings when he was only a six grade student at elementry school.

This summer he was working at Tencent as an intern. One day his leader came to ask xiaoxin for help. His leader gave him a string and he wanted xiaoxin to generate palindromic strings for him. Once xiaoxin generates a different palindromic string, his leader will give him a watermelon candy. The problem is how many candies xiaoxin‘s leader needs to buy?

 

 

Input

This problem has multi test cases. First line contains a single integer T(T≤20) which represents the number of test cases.
For each test case, there is a single line containing a string S(1≤length(S)≤1,000) .

 

 

Output

For each test case, print an integer which is the number of watermelon candies xiaoxin‘s leader needs to buy after mod 1,000,000,007 .

 

 

Sample Input

3

aa

aabb

a

 

 

Sample Output

1

2

1

 

 

Source

BestCoder Round #77 (div.2)

 

 

Recommend

wange2014   |   We have carefully selected several similar problems for you:  5659 5658 5657 5654 5653 

 

在网上找了模板，虽然我没有看懂，但是直接用了就过了。。。

 

题意：给你一个字符串，可以打乱顺序，问可以组合成几种回文串。

 

附上代码：

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 const int mod=1e9+7;
 7 typedef long long ll;
 8 //返回d=gcd(a,b);和对应于等式ax+by=d中的x,y
 9 ll extend_gcd(ll a,ll b,ll &x,ll &y)
10 {
11     if(a==0&&b==0) return -1;//无最大公约数
12     if(b==0)
13     {
14         x=1;
15         y=0;
16         return a;
17     }
18     ll d=extend_gcd(b,a%b,y,x);
19     y-=a/b*x;
20     return d;
21 }
22 //*********求逆元素*******************
23 //ax = 1(mod n)
24 ll mod_reverse(ll a,ll n)
25 {
26     ll x,y;
27     ll d=extend_gcd(a,n,x,y);
28     if(d==1) return (x%n+n)%n;
29     else return -1;
30 }
31 
32 ll c(ll m,ll n)
33 {
34     ll i,j,t1,t2,ans;
35     t1=t2=1;
36     for(i=n; i>=n-m+1; i--) t1=t1*i%mod;
37     for(i=1; i<=m; i++) t2=t2*i%mod;
38     return  t1*mod_reverse(t2,mod)%mod;
39 }
40 
41 int main()
42 {
43     int T,i,j,n,m;
44     char ch[1005];
45     int a[30];
46     scanf("%d",&n);
47     while(n--)
48     {
49         memset(a,0,sizeof(a));
50         scanf("%s",ch);
51         for(i=0; ch[i]!=‘\0‘; i++)
52             a[ch[i]-‘a‘]++;
53         int t=0;
54         for(i=0; i<26; i++)
55         {
56             if(a[i]%2!=0)
57                 t++;
58             if(t>1)
59                 break;
60         }
61         if(i!=26)
62         {
63             printf("0\n");
64             continue;
65         }
66         int len=strlen(ch);
67         ll sum=1;
68         len/=2;
69         for(i=0; i<26; i++)
70         {
71             sum=(sum*c(a[i]/2,len))%mod;
72             len-=a[i]/2;
73         }
74         printf("%d\n",sum);
75     }
76     return 0;
77 }