亚洲免费在线-亚洲免费在线播放-亚洲免费在线观看-亚洲免费在线观看视频-亚洲免费在线看-亚洲免费在线视频

Password Attacker

系統(tǒng) 1941 0

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!

Password Attacker


更多文章、技術(shù)交流、商務(wù)合作、聯(lián)系博主

微信掃碼或搜索:z360901061

微信掃一掃加我為好友

QQ號聯(lián)系: 360901061

您的支持是博主寫作最大的動力,如果您喜歡我的文章,感覺我的文章對您有幫助,請用微信掃描下面二維碼支持博主2元、5元、10元、20元等您想捐的金額吧,狠狠點擊下面給點支持吧,站長非常感激您!手機微信長按不能支付解決辦法:請將微信支付二維碼保存到相冊,切換到微信,然后點擊微信右上角掃一掃功能,選擇支付二維碼完成支付。

【本文對您有幫助就好】

您的支持是博主寫作最大的動力,如果您喜歡我的文章,感覺我的文章對您有幫助,請用微信掃描上面二維碼支持博主2元、5元、10元、自定義金額等您想捐的金額吧,站長會非常 感謝您的哦!!!

發(fā)表我的評論
最新評論 總共0條評論
主站蜘蛛池模板: 午夜免费体验区 | 久久久精品视频免费观看 | 欧美精品成人一区二区视频一 | 亚洲精品中文字幕一区 | 久草精品视频在线播放 | 免费国产成人高清在线观看麻豆 | 午夜久久久 | 国产精品伦理 | 亚洲欧美香蕉在线日韩精选 | 久久66热re国产毛片基地 | 99九九精品免费视频观看 | 国产欧美曰韩一区二区三区 | 国产资源福利 | 亚洲成人精品视频 | 牛人盗摄一区二区三区视频 | 青青青手机版视频在线观看 | 亚洲日韩精品欧美一区二区 | 全午夜免费一级毛片 | 激情婷婷成人亚洲综合 | 九九热精品视频在线播放 | 国产视频二区在线观看 | 久久综合精品不卡一区二区 | 亚洲一区二区三区欧美 | 四虎四虎1515hhcom | 久久中文字幕亚洲精品最新 | 国产亚洲精品美女久久久 | 色婷婷亚洲十月十月色天 | 全部免费的毛片在线看青青 | 久久这里只有精品免费视频 | 香香在线观看视频 | sihu国产精品永久免费 | 欧美不卡在线视频 | 黄页网址大全免费观看美女 | 女人十八毛片免费特黄 | 欧美精品久久久久久久久大尺度 | 国产精彩视频 | aaa一级毛片 | 日本高清视频一区二区 | 免费视频一区 | 天天综合日日噜噜噜 | 国产午夜亚洲精品久久999 |