Problem A
Algebraic Teamwork
The great pioneers of group theory and linear algebra want to cooperate and join their theories. In group theory, permutations – also known as bijective functions – play an important role. For a finite set A, a function σ : A → A is called a permutation of A if and only if there is some function ρ : A → A with σ(ρ(a)) = a and ρ(σ(a)) = a for all a ∈ A.
The other half of the new team – the experts on linear algebra – deal a lot with idempotent functions. They appear as projections when computing shadows in 3D games or as closure operators like the transitive closure, just to name a few examples. A function p : A → A is called idempotent if and only if p(p(a)) = p(a) for all a ∈ A.
To continue with their joined research, they need your help. The team is interested in non-idempotent permutations of a given finite set A. As a first step, they discovered that the result only depends on the set’s size. For a concrete size 1 ≤ n ≤ 105, they want you to compute the number of permutations on a set of cardinality n that are not idempotent.
Input
The input starts with the number t ≤ 100 of test cases. Then t lines follow, each containing the set’s size 1 ≤ n ≤ 105.
Output
Output one line for every test case containing the number modulo 1000000007 = (109 + 7) of non-idempotent permutations on a set of cardinality n.
|
Sample Input |
Sample Output |
|
3 1 2 2171 |
0 1 6425 |
解题：n!-1
1 #include <bits/stdc++.h> 2 #define LL long long 3 using namespace std; 4 const int maxn = 100010; 5 const int md = 1e9+7; 6 LL d[maxn]; 7 int main(){ 8 d[0] = 1; 9 for(int i = 1; i < maxn; ++i) 10 d[i] = (i%md*d[i-1])%md; 11 int x,ks; 12 scanf("%d",&ks); 13 while(ks--){ 14 scanf("%d",&x); 15 printf("%lld ",(d[x]+md-1)%md); 16 } 17 return 0; 18 }