Problem A. Password Attacker

Problem A. Password Attacker

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Small input
8 points

Solve A-small

Judge's response for last submission: Correct.

Large input
13 points

Solve A-large

Judge's response for last submission: Correct.

Problem

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 Mkeys. 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 
 
Output 
 

41 13 45 515 15

Case #1: 1Case #2: 36Case #3: 120Case #4: 674358851

第一次看到自己写的代码比别人的短，而且思路直接，纪念一下。
题意：m个不同字母，填充长度为n的字符串，m个字符必须都用上，不够的部分可以用这m个字符中的任意字符重复填充，问可能有多少种
字符串。
思路：开始想用排列组合一步搞定，发现不行，然后就想到可以递推解决，对并不是动态规划。
递推状态如下定义：f[i][j]表示长度为 i的字符串，用j种字符填充的方法数，
递推关系如下：
f[i][j] = f[i-1][j]×j + f[i-1][j-1]×j


f[i-1][j]×j表示在长度为i-1的字符串后面再加一个位置，那么这个位置可以选j种字符中的任意一种填充。
所以这样得到的字符串有f[i-1][j]×j种，


f[i-1][j-1]×j 表示长度为i-1的字符串再添一个位置，这个位置需要用之前没有用过的颜色，就是j种
颜色里面选一种，有j种选法所以要 乘 j
代码如下：

#include <bits/stdc++.h>const int N = 110;typedef long long ll;const int mod = 1e9 + 7;using namespace std;ll f[N][N];void run(){  int n, m;  scanf("%d%d", &m, &n);  printf("%lld\n", f[n][m]);}int main(){#ifndef ONLINE_JUDGE      freopen("in.txt","r" ,stdin);#endif     int T, cas = 1;    for (int i = 1; i <= 100; i++)      f[i][1] = 1;    for (int i = 2; i <= 100; i++)      for (int j = 1; j <= 100; j++)      {        f[i][j] = f[i-1][j] * j % mod;        f[i][j] += f[i-1][j-1] * j %mod;        f[i][j] %= mod;      }    scanf("%d", &T);    while (T--)    {      printf("Case #%d: ", cas++);      run();    }    return 0; }