Topic links: https://atcoder.jp/contests/abc127/tasks/abc127_e
Subject to the effect
Given a $ N * M $ board, tuple $ (x, y), 1 \ leq x \ leq N, 1 \ leq y \ leq M $, represents a certain position on the board, the board now K is selected from the different positions, referred to as $ (x_1, y_1), (x_2, y_2), \ dots, (x_K, y_K) corresponding cost of $, selected for $ \ sum_ {i = 1} ^ {K- 1} \ sum_ {j = i + 1} ^ K $ sum of the cost, all possible output schemes (| x_i - | - x_j | | + y_j y_i).
analysis
First difficult to find, x and y can be calculated separately, the only requirement $ \ sum_ {i = 1} ^ {K-1} \ sum_ {j = i + 1} ^ K | x_i - x_j | $ i.e. can, similarly calculated $ \ sum_ {i = 1} ^ {K-1} \ sum_ {j = i + 1} ^ K | y_i - y_j | $.
If we fix the position of the board 2 $ x_ {i_1, j_1}, x_ {i_2, j_2} $ does not move, in this case, $ \ tbinom {N * M-2} {K-2} $ species option, in other words $ | x_ {i_1, j_1} - x_ {i_2, j_2} | $ appeared $ \ tbinom {N * M-2} {K-2} $ times.
But to enumerate all the points undoubtedly to overtime, for which we can enumerate $ d = | x_ {i_1, j_1} - x_ {i_2, j_2} | $, $ d \ in [1, K - 1] $ .
Law can first look, when d == 1, the number of look $ (x_ {i_1, j_1}, x_ {i_2, j_2}) $ is satisfied.
It can be found as long as the difference between the two points is a vertical axis, they all satisfied.
So when d == 1 when there is $ M ^ 2 * (N - 1) $ species $ (x_ {i_1, j_1}, x_ {i_2, j_2}) $ satisfies $ d == | x_ {i_1, j_1} - x_ {i_2, j_2} | $.
And so on, can find out the rules: for each d, has $ d * M ^ 2 * (N - d) $ species $ (x_ {i_1, j_1}, x_ {i_2, j_2}) $ $ d satisfies == | x_ {i_1, j_1} - x_ {i_2, j_2} | $.
Then for each d, its contribution to the answer is $ \ tbinom {N * M-2} {K-2} * d * M ^ 2 * (N - d) $.
All add up to.
code show as below
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); 5 #define Rep(i,n) for (int i = 0; i < (n); ++i) 6 #define For(i,s,t) for (int i = (s); i <= (t); ++i) 7 #define rFor(i,t,s) for (int i = (t); i >= (s); --i) 8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i) 9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i) 10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i) 11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i) 12 13 #define pr(x) cout << #x << " = " << x << " " 14 #define prln(x) cout << #x << " = " << x << endl 15 16 #define LOWBIT(x) ((x)&(-x)) 17 18 #define ALL(x) x.begin(),x.end() 19 #define INS(x) inserter(x,x.begin()) 20 21 #define ms0(a) memset(a,0,sizeof(a)) 22 #define msI(a) memset(a,inf,sizeof(a)) 23 #define msM(a) memset(a,-1,sizeof(a)) 24 25 #define MP make_pair 26 #define PB push_back 27 #define ft first 28 #define sd second 29 30 template<typename T1, typename T2> 31 istream &operator>>(istream &in, pair<T1, T2> &p) { 32 in >> p.first >> p.second; 33 return in; 34 } 35 36 template<typename T> 37 istream &operator>>(istream &in, vector<T> &v) { 38 for (auto &x: v) 39 in >> x; 40 return in; 41 } 42 43 template<typename T1, typename T2> 44 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) { 45 out << "[" << p.first << ", " << p.second << "]" << "\n"; 46 return out; 47 } 48 49 inline int gc(){ 50 static const int BUF = 1e7; 51 static char buf[BUF], *bg = buf + BUF, *ed = bg; 52 53 if(bg == ed) fread(bg = buf, 1, BUF, stdin); 54 return *bg++; 55 } 56 57 inline int ri(){ 58 int x = 0, f = 1, c = gc(); 59 for(; c<48||c>57; f = c=='-'?-1:f, c=gc()); 60 for(; c>47&&c<58; x = x*10 + c - 48, c=gc()); 61 return x*f; 62 } 63 64 typedef long long LL; 65 typedef unsigned long long uLL; 66 typedef pair< double, double > PDD; 67typedef pair < int , int > PII; 68 typedef pair < string , int > PSI; 69 typedef Set < int > SI; 70 typedef vector < int > VI; 71 typedef vector <PII> VPII; 72 typedef map < int , int > MII; 73 typedef pair <LL, LL> PLL; 74 typedef vector <LL> VL; 75 typedef vector <VL> VVL; 76 const double EPS = 1e-10; 77 const LL inf = 0x7fffffff; 78 const LL infLL = 0x7fffffffffffffffLL; 79 const LL mod = 1e9 + 7; 80 const int maxN = 2e5 + 7; 81 const LL ONE = 1; 82 const LL evenBits = 0xaaaaaaaaaaaaaaaa; 83 const LL oddBits = 0x5555555555555555; 84 85 LL fac[maxN]; 86 void init_fact () { 87 FAC [ 0 ] = . 1 ; 88 the For (I, . 1 , maxN - . 1 ) { 89 FAC [I] = (I * FAC [I - . 1 ])% MOD; 90 } 91 is } 92 93 // AX + by = GCD (A, B) = D 94 // extended Euclidean algorithm 95 / * * 96 * A + B * X * Y. 1 = 97 * ab If prime, solvability 98 * x b is the inverse element on a 99 * Y b is the inverse of the element on a 100 * 101 * proof: 102 * A * X% B + B * Y% B =. 1% B 103 * A * X% B =. 1% B 104 * A * X =. 1 (MOD B) 105 * / 106 inline void ex_gcd (LL A, LL B, LL & X, LL & Y, LL & D) { 107 IF (B!) {D = A, X = . 1 , Y = 0 ;} 108 the else { 109 ex_gcd (B, A% B, Y, X, D); 110 Y - X * = (a / B); 111 } 112 } 113 114 // find the inverse element on a p, and if there is, return -1 115 // a与p互质，逆元才存在 116 inline LL inv_mod(LL a, LL p = mod){ 117 LL d, x, y; 118 ex_gcd(a, p, x, y, d); 119 return d == 1 ? (x % p + p) % p : -1; 120 } 121 122 inline LL comb_mod(LL m, LL n) { 123 LL ret; 124 if(m > n) swap(m, n); 125 ret = (fac[n] * inv_mod(fac[m], mod)) % mod; 126 ret = (ret * inv_mod(fac[n - m], mod)) % mod; 127 return ret; 128 } 129 130 void add_mod(LL &a, LL b) { 131 a = (a + b) % mod; 132 if(a < 0) a += mod; 133 } 134 135 int N, M, K; 136 LL ans; 137 138 int main(){ 139 INIT(); 140 init_fact(); 141 cin >> N >> M >> K; 142 LL cnt = comb_mod(K - 2, N * M - 2); 143 144 ForLL(d, 1, N - 1) add_mod(ans, cnt * ((d * M * M * (N - d)) % mod)); 145 ForLL(d, 1, M - 1) add_mod(ans, cnt * ((d * N * N * (M - d)) % mod)); 146 cout << ans << endl; 147 return 0; 148 }