UVa 11481 Arrange the Numbers (组合数学)

时间:2022-01-12 08:07:22

题意:给定 n,m,k,问你在 1 ~ n 的排列中,前 m 个恰好有 k 个不在自己位置的排列有多少个。

析:枚举 m+1 ~ n 中有多少个恰好在自己位置,这个是C(n-m, i),然后前面选出 k 个,是C(m, k),剩下 n - k - i 个是都不在自己位置,也就是错排 D[n-k-i],求一个和就Ok了。

代码如下:

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define be begin()
#define ed end()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
//#define aLL 1,n,1
#define FOR(i,n,x) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.in", "r", stdin)
#define freopenw freopen("out.out", "w", stdout)
using namespace std; typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int maxn = 1000 + 10;
const int maxm = 76543;
const int mod = 1000000007;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
inline int readInt(){ int x; scanf("%d", &x); return x; } int f[maxn], D[maxn], inv[maxn]; int fast_pow(int a, int n){
int res = 1;
while(n){
if(n&1) res = (LL)res * a % mod;
n >>= 1;
a = (LL)a * a % mod;
}
return res;
} int main(){
D[1] = 0; D[0] = D[2] = 1;
f[0] = f[1] = 1; f[2] = 2;
for(int i = 3; i <= 1000; ++i){
f[i] = (LL)f[i-1] * i % mod;
D[i] = (LL)(i-1) * (D[i-1] + D[i-2]) % mod;
}
inv[1000] = fast_pow(f[1000], mod-2);
for(int i = 999; i >= 0; --i)
inv[i] = (LL)(i+1) * inv[i+1] % mod;
int T, k; cin >> T;
for(int kase = 1; kase <= T; ++kase){
scanf("%d %d %d", &n, &m, &k);
int ans = 0;
int cmk = (LL)f[m] * inv[k] % mod * inv[m-k] % mod;
for(int i = 0; i <= n-m; ++i)
ans = (ans + (LL)cmk * f[n-m] % mod * inv[i] % mod * inv[n-m-i] % mod * D[n-k-i] % mod) % mod;
printf("Case %d: %d\n", kase, ans);
}
return 0;
}