题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6143
题意：给m个字符，要求分配m个字符给许多二元组，每个二元组包含两个字符串，长度为n。要求两个字符串中不能出现相同的字符，问有多少种二元组。
考虑给二元组中第一个字符串i个字符，那么有C(m,i)种选法。希望把这i个字符都用上。那么就用dp来处理。
dp(i,j)表示字符串长为i，用j个字符凑成，有多少种凑法。考虑当前字符用完后下次还用或者不用，一共有j种选法，就是dp(i,j)=j*(dp(i-1,j)+dp(i-1,j-1))。记搜下就行，注意j>i的时候返回0。
右边的字符串就不用管，字符可以有不用上的，就是(m-i)^n种排列方式。
1 /* 2 ━━━━━┒ギリギリ♂ eye！ 3 ┓┏┓┏┓┃キリキリ♂ mind！ 4 ┛┗┛┗┛┃＼○／ 5 ┓┏┓┏┓┃ / 6 ┛┗┛┗┛┃ノ) 7 ┓┏┓┏┓┃ 8 ┛┗┛┗┛┃ 9 ┓┏┓┏┓┃ 10 ┛┗┛┗┛┃ 11 ┓┏┓┏┓┃ 12 ┛┗┛┗┛┃ 13 ┓┏┓┏┓┃ 14 ┃┃┃┃┃┃ 15 ┻┻┻┻┻┻ 16 */ 17 #include <bits/stdc++.h> 18 using namespace std; 19 #define fr first 20 #define sc second 21 #define cl clear 22 #define BUG puts("here!!!") 23 #define W(a) while(a--) 24 #define pb(a) push_back(a) 25 #define Rint(a) scanf("%d", &a) 26 #define Rll(a) scanf("%I64d", &a) 27 #define Rs(a) scanf("%s", a) 28 #define Cin(a) cin >> a 29 #define FRead() freopen("in", "r", stdin) 30 #define FWrite() freopen("out", "w", stdout) 31 #define Rep(i, len) for(int i = 0; i < (len); i++) 32 #define For(i, a, len) for(int i = (a); i < (len); i++) 33 #define Cls(a) memset((a), 0, sizeof(a)) 34 #define Clr(a, x) memset((a), (x), sizeof(a)) 35 #define Full(a) memset((a), 0x7f7f7f, sizeof(a)) 36 #define lrt rt << 1 37 #define rrt rt << 1 | 1 38 #define pi 3.14159265359 39 #define RT return 40 #define lowbit(x) x & (-x) 41 #define onenum(x) __builtin_popcount(x) 42 typedef long long LL; 43 typedef long double LD; 44 typedef unsigned long long ULL; 45 typedef pair<int, int> pii; 46 typedef pair<string, int> psi; 47 typedef pair<LL, LL> pll; 48 typedef map<string, int> msi; 49 typedef vector<int> vi; 50 typedef vector<LL> vl; 51 typedef vector<vl> vvl; 52 typedef vector<bool> vb; 53 54 const LL mod =(LL) 1e9+7; 55 const int maxn = 2100001; 56 LL f[maxn], dp[2020][2020]; 57 int n, m; 58 59 void init() { 60 memset(dp, -1, sizeof(dp)); 61 f[0] = f[1] = 1; 62 for(int i = 2; i < maxn; i++) f[i] = (f[i-1] * i) % mod; 63 } 64 LL mul(LL x, LL n) { 65 LL ret = 1; 66 while(n) { 67 if(n & 1) ret = (ret * x) % mod; 68 n >>= 1; 69 x = (x * x) % mod; 70 } 71 return ret; 72 } 73 LL exgcd(LL a, LL b, LL &x, LL &y) { 74 if(b == 0) { 75 x = 1; 76 y = 0; 77 return a; 78 } 79 else { 80 LL ret = exgcd(b, a%b, x, y); 81 LL tmp = x; 82 x = y; 83 y = tmp - a / b * y; 84 return ret; 85 } 86 } 87 LL inv(LL a) { 88 LL x, y; 89 exgcd(a, mod, x, y); 90 return (x % mod + mod) % mod; 91 } 92 LL C(LL x, LL y) { 93 return f[x] * inv(f[x-y]) % mod * inv(f[y]) % mod; 94 } 95 96 LL DP(int n, int m) { 97 if(m > n) return 0; 98 if(dp[n][m] != -1) return dp[n][m]; 99 if(n == m) return dp[n][m] = f[n]; 100 if(m == 1) return dp[n][m] = 1; 101 return dp[n][m] = m * (DP(n-1, m) + DP(n-1, m-1) % mod) % mod; 102 } 103 104 signed main() { 105 // FRead(); 106 int T; 107 Rint(T); 108 init(); 109 W(T) { 110 Rint(n); Rint(m); 111 LL ret = 0; 112 For(i, 1, m) { 113 LL left = 0, right = 0; 114 ret += C(m, i) * DP(n, i) % mod * mul(m-i, n) % mod; 115 ret %= mod; 116 } 117 printf("%I64d ", ret); 118 } 119 RT 0; 120 }