BZOJ 3434 Wc2014 time-space shuttle Mobius Inversion, bzoj3434

BZOJ 3434 Wc2014 time-space shuttle Mobius Inversion, bzoj3434

For a given n-dimensional space, we need to select c collocated points in the n-dimensional space, which requires that the coordinates of these points increase monotonically for each dimension, the I-dimension coordinate is an integer in [1, mi]

In order to express convenience | the number of solutions that represent the intermediate thing. By the way, the I-dimensional coordinates are set to ai, but this is not concise. Therefore, x is used to represent the 1st-dimensional coordinates, and y is used to represent the second-dimensional coordinates, θ indicates the n-dimensional coordinate

Seems like my method is SB? No matter whether it's good to push it out --

Nima BZOJ slag evaluation machine card constant--clearly said that UOJ5s has passed --

# Include <cstdio> # include <cstring> # include <iostream> # include <algorithm> # define M 100100 # define MOD 10007 using namespace std; int mu [M], prime [M], tot; bool not_prime [M]; int f [M] [20], mu_f [M] [20], sum [M] [20] [12]; // f [I] [j] indicates the sum of I, number of positive integers with a length of j // mu_f [T] [j] indicates the value of Σ [I | T] mu [I] F [T/I] When C-1 = j // sum [T] [j] [k] indicates C-1 fix and int n, c, min_m, m [12], a [12]; void Linear_Shaker () {int I, j; mu [1] = 1; for (I = 2; I <= 10000 0; I ++) {if (! Not_prime [I]) {prime [++ tot] = I; mu [I] =-1 ;}for (j = 1; prime [j] * I <= 100000; j ++) {not_prime [prime [j] * I] = 1; if (I % prime [j] = 0) {mu [prime [j] * I] = 0; break;} mu [prime [j] * I] =-mu [I] ;}} void Pretreatment () {int I, j, k; Linear_Shaker (); for (I = 1; I <= 100000; I ++) {f [I] [1] = 1; for (j = 2; j <= I & j <= 19; j ++) f [I] [j] = (f [I-1] [j] + f [I-1] [J-1]) % MOD;} for (I = 1; I <= 100000; I ++) for (j = 1; j <= 19; j ++) for (k = 100000/I; k --) (mu_f [k * I] [j] + = mu [I] * f [k] [j]) % = MOD; for (I = 1; I <= 100000; I ++) for (j = 1; j <= 19; j ++) {int temp = 1; for (k = 0; k <= 11; k ++, temp * = I, temp % = MOD) (sum [I] [j] [k] = sum [I-1] [j] [k] + mu_f [I] [j] * temp % MOD) % = MOD ;}} void Get_A (int lower) {int I, j; memset (a, 0, sizeof a); a [0] = 1; for (I = 1; I <= n; I ++) {int d = m [I]/lower; int k =-(long) d * (d + 1)> 1) % MOD; int B = (long) m [I] * d) % MOD; for (j = n; j; j --) (a [j] = k * a [J-1] + B * a [j]) % = MOD; (a [0] * = B) % = MOD ;}} int main () {int T, I, j, last; Pretreatment (); for (cin> T; T --) {scanf ("% d", & n, & c); min_m = 0x3f3f3f; for (I = 1; I <= n; I ++) {scanf ("% d", & m [I]); min_m = min (min_m, m [I]);} int ans = 0; for (I = 1; I <= min_m; I = last + 1) {last = 0x3f3f3f3f; for (j = 1; j <= n; j ++) last = min (last, m [j]/(m [j]/I); Get_A (I); for (j = 0; j <= n; j ++) (ans + = a [j] * (sum [last] [C-1] [j]-sum [I-1] [C-1] [j]) % MOD) % = MOD ;} printf ("% d \ n", (ans + MOD) % MOD);} return 0 ;}