Passwords are widely used in our lives: for ATMs, online forum logins, mobile device unlock and door access. Everyone cares about password security. However, attackers always find ways to steal our passwords. Here is one possible situation:
Assume that Eve, the attacker, wants to steal a password from the victim Alice. Eve cleans up the keyboard beforehand. After Alice types the password and leaves, Eve collects the fingerprints on the keyboard. Now she knows which keys are used in the password. However, Eve won't know how many times each key has been pressed or the order of the keystroke sequence.
To simplify the problem, let's assume that Eve finds Alice's fingerprints only occurs on M keys. And she knows, by another method, that Alice's password contains N characters. Furthermore, every keystroke on the keyboard only generates a single, unique character. Also, Alice won't press other irrelevant keys like 'left', 'home', 'backspace' and etc.
Here's an example. Assume that Eve finds Alice's fingerprints on M=3 key '3', '7' and '5', and she knows that Alice's password is N=4-digit in length. So all the following passwords are possible: 3577, 3557, 7353 and 5735. (And, in fact, there are 32 more possible passwords.)
However, these passwords are not possible:
1357 // There is no fingerprint on key '1'
3355 // There is fingerprint on key '7',
so '7' must occur at least once.
357 // Eve knows the password must be a 4-digit number.
With the information, please count that how many possible passwords satisfy the statements above. Since the result could be large, please output the answer modulo 1000000007(109+7).
Input
The first line of the input gives the number of test cases, T.
For the next T lines, each contains two space-separated numbers M and N, indicating a test case.
Output
For each test case, output one line containing "Case #x: y", where x is the test case number (starting from 1) and y is the total number of possible passwords modulo 1000000007(109+7).
Limits
Small dataset
T = 15.
1 ≤ M ≤ N ≤ 7.
Large dataset
T = 100.
1 ≤ M ≤ N ≤ 100.
Sample
Input
4
1 1
3 4
5 5
15 15
Output?
Case #1: 1
Case #2: 36
Case #3: 120
Case #4: 674358851
google在線筆試題。這題一直沒做出來。有人說有公式，看到大牛們提交的代碼又覺得像是dp。后來學(xué)了生成函數(shù)之后，覺得應(yīng)該是一道指數(shù)型生成函數(shù)的題。
1
#include <iostream>
2
#include <cstdio>
3
#include <vector>
4
5
using
namespace
std;
6
const
double
epi =
0.000001
;
7
int
frac(
int
k) {
8
int
ans =
1
;
9
for
(
int
i =
2
; i <= k; ++
i) {
10
ans *=
i;
11
}
12
return
ans;
13
}
14
15
int
enumPassword(
int
n,
int
m) {
16
vector<vector<
double
> >
params
(
2
, vector<
double
>(n +
1
,
0
));
17
params
[
0
][
0
] =
1
;
18
int
cur =
0
, next =
1
;
19
20
for
(
int
i =
0
; i < m; ++
i) {
21
params
[next].assign(n +
1
,
0
);
22
for
(
int
j =
0
; j <= n; ++
j) {
23
if
(
params
[cur][j] < epi)
continue
;
24
for
(
int
k =
1
; k + j <= n; ++
k) {
25
params
[next][k + j] =
params
[next][k + j] +
params
[cur][j] *
1
/
frac(k);
26
}
27
}
28
cur = !cur; next = !
next;
29
}
30
31
return
params
[cur][n] *
frac(n);
32
}
33
34
int
main(
int
argc,
char
**
argv) {
35
if
(argc <
2
)
return
-
1
;
36
freopen(argv[
1
],
"
r
"
, stdin);
37
int
test;
38
scanf(
"
%d
"
, &
test);
39
for
(
int
i =
0
; i < test; ++
i) {
40
int
m, n;
41
scanf(
"
%d%d
"
, &m, &
n);
42
cout <<
"
Case #
"
<< i +
1
<<
"
:
"
<< enumPassword(n, m) <<
endl;
43
}
44
45
return
0
;
46
}
但是數(shù)太大，要取模。除操作不能直接除模。
在網(wǎng)上搜到一個定理：
定理12.2：設(shè)$a_n$,$b_n$的指數(shù)生成函數(shù)分別為f(x)和g(x)，則：
$f(x)*g(x) = \sum_{n=0}^{\infty}c_n\frac{x^n}{n!}, c_n = \sum_{k=0}^{n}C(n,k)a_kb_{n-k}$。
對應(yīng)到我們這里，Line 25里就變成params[cur][j]*1*C(j+k, k)，params[cur][j]對應(yīng)的是$\frac{x^j}{j!}$的系數(shù)，1對應(yīng)的是$\frac{x^k}{k!}$，所以乘以的就是C(j+k, k)了。
代碼如下：
1
#include <iostream>
2
#include <cstdio>
3
#include <vector>
4
5
using
namespace
std;
6
enum
{MOD =
1000000007
};
7
typedef
long
long
llong;
8
llong combination[
100
][
100
];
9
void
getCombination() {
10
for
(
int
i =
0
; i <=
100
; ++
i) {
11
for
(
int
j =
0
; j <= i; ++
j) {
12
if
(j ==
0
) {
13
combination[i][j] =
1
;
14
}
else
{
15
combination[i][j] = (combination[i -
1
][j] + combination[i -
1
][j -
1
]) %
MOD;
16
}
17
}
18
}
19
}
20
21
llong enumPassword(
int
n,
int
m) {
22
vector<vector<llong> >
params
(
2
, vector<llong>(n +
1
,
0
));
23
params
[
0
][
0
] =
1
;
24
int
cur =
0
, next =
1
;
25
26
for
(
int
i =
0
; i < m; ++
i) {
27
params
[next].assign(n +
1
,
0
);
28
for
(
int
j =
0
; j <= n; ++
j) {
29
if
(
params
[cur][j] ==
0
)
continue
;
30
for
(
int
k =
1
; k + j <= n; ++
k) {
31
params
[next][k + j] = (
params
[next][k + j] +
params
[cur][j] * combination[j + k][k]) %
MOD;
32
}
33
}
34
cur = !cur; next = !
next;
35
}
36
37
return
params
[cur][n];
38
}
39
40
int
main(
int
argc,
char
**
argv) {
41
if
(argc <
2
)
return
-
1
;
42
freopen(argv[
1
],
"
r
"
, stdin);
43
if
(argc >=
3
) freopen(argv[
2
],
"
w
"
, stdout);
44
getCombination();
45
int
test;
46
scanf(
"
%d
"
, &
test);
47
for
(
int
i =
0
; i < test; ++
i) {
48
int
m, n;
49
scanf(
"
%d%d
"
, &m, &
n);
50
//
cout << "Case #" << i + 1 << ": " << enumPassword(n, m) << endl;
51
printf(
"
Case #%d: %lld\n
"
, i +
1
, enumPassword(n, m));
52
}
53
54
return
0
;
55
}
注意這里求組合數(shù)要用遞推公式來求，這樣可以在運算中取模，避免溢出。
$C(n, m) = C(n - 1, m) + C(n - 1, m - 1)$。
以后指數(shù)型生成函數(shù)的題都可以這么做。get！
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