codechef 营养题 第一弹

第一弾が始まる！

定期更新しない！

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codechef problems 第一弹

一.Authentication Failed
原题题面
Several days ago Chef decided to register at one of the

programming sites. For registering he was asked to choose a

nickname and a password. There was no problem with choosing a

nickname ("Chef" is his favorite nickname), but choosing a

password in a secure way seemed to be a real problem for Chef.

Therefore, he decided to write a program which would generate

the password of length N consisting of small Latin letters a..z.

Then Chef successfully registered at the site and saved the

password in a file (as it was too hard to remember).

Today Chef decided to visit the site once again. He entered his

nickname, copied the password from the file... "Authentication

failed!" was the answer. Trying to understand the reason of

this, he noticed that the password in his file had length N+K

instead of N! Sure enough of the source of the problem, Chef

went straight to his young brother.

And Chef was right, it was his brother who had inserted K random

small Latin letters at some random positions (possibly at the

beginning or at the end) of the password. Chef's brother didn't

remember what exactly changes he had made at all, but he

promised that he had done nothing besides inserting letters.

As there is no other way to recover the password, Chef is now

starting to remove every possible combination of K letters from

the password trying to enter (when Chef obtains the same

password as one of the previously entered passwords, he doesn't

try to enter using this password again). Now the question is:

what is the number of times Chef will receive "Authentication

failed!" as the answer before successful entering in the worst

case? As the answer might be quite large, output its remainder

of division by 1009419529.

Input

The first line of the input file contains one integer T -- the

number of test cases (no more than 10). Each test case is

described by a line containing two integers N and K (1 ≤ N ≤

10000, 1 ≤ K ≤ 100) separated by a single space, followed by a

line containing a string of length N+K consisting of small Latin

letters a..z.

Output

For each test case output just one line containing the required

number modulo 1009419529.

Example

Input:
3
2 1
aaa
3 1
abcd
4 2
ababab

Output:
0
3
10

Explanation:
In the first test case, the password is definitely "aa". In the

second test case, it can be "abc", "abd", "acd" or "bcd", so in

the worst case Chef will guess the correct option from the

fourth attempt, thus making 3 unsuccessful attempts.

Description
你把一个长为N的小写字母组成的密码保存在一个txt文件里；一个熊孩子

在密码的某些位置插入了共计K个字母，注意这里的K个字母不存在重复；

你决定把密码重新试出来；求最坏的情况下需要试多少次？
规定：N<=1W,K<=100

Solution
我们选择用递推的方式计数
先设计计数状态吧
设f[i][j]为前i位中去掉了j位且第i个字符未删除的方案数
那么我们的递推式是f[i][j]=f[i-1][j]+f[i-2][j-1]+f[i-3][j-2]+...
直接枚举每次累加的时间复杂度为O(N*K*K)，超时
由于除了f[i-1][j]以外，其他项可以看作是递推矩阵中一个对角线上的

数字和，即∑f[i-x][j-x+1]
那么考虑前缀和优化，多加几个数组就行了
优化后的时间复杂度为O(N*K)，O(100W)->AC
Code

#include <stdio.h>
#include <memory.h>
#define MaxL 10110
#define MaxK 110
#define MaxBuf 1<<22
#define mo 1009419529
#define RG register
#define Blue() ((S==T&&(T=(S=B)+fread(B,1,MaxBuf,stdin),S==T))?0:*S++)
#define dmin(a,b) ((a)<(b)?(a):(b))
#define dmax(a,b) ((a)>(b)?(a):(b))
char B[MaxBuf],*S=B,*T=B;
template<class Type>inline void Rin(RG Type &x){
    x=;RG int c=Blue();
    for(;c<||c>;c=Blue())
        ;
    for(;c>&&c<;c=Blue())
        x=(x<<)+(x<<)+c-;
}
inline void geTc(char *C){
    char c=Blue();
    for(;c<'a'||c>'z';c=Blue())
        ;
    for(;c>='a'&&c<='z';c=Blue())
        *C++=c;
}
char ch[MaxL];
int kase,n,k,f[MaxL][MaxK],g[MaxL],h[][MaxL],ans;
#define FO(x) {freopen(#x".in","r",stdin);}
int main(){
    FO(cc authen failed);
    Rin(kase);
    while(kase--){
        memset(f,,sizeof f);
        memset(g,,sizeof g);
        memset(h,,sizeof h);
        ans=;
        Rin(n),Rin(k),geTc(ch);
        f[][]=g[]=;
        for(RG int i=;i<=n+k;i++){
            RG int s=ch[i-]-'a';
            for(RG int j=;j<=dmin(i-,k);j++){
                f[i][j]=(g[i-j-]-h[s][i-j])%mo;
                if(i-j==n)(ans+=f[i][j])%=mo;
                (g[i-j]+=f[i][j])%=mo;
                (h[s][i-j]+=f[i][j])%=mo;
            }
        }
        printf("%d\n",(ans-+mo)%mo);
    }
    return ;
}