Let Me Count The Ways(Kickstart Round H 2018)

题目链接：https://code.google.com/codejam/contest/3324486/dashboard#s=p2

题目：

思路：

　　

代码实现如下：

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pli;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;

#define lson rt<<1
#define rson rt<<1|1
#define lowbit(x) x&(-x)
#define  name2str(name) (#name)
#define bug printf("*********
")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("D://code//in.txt", "r", stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)

const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 2e5 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;

int t, n, m;
LL A[maxn], inv[maxn];

LL Mod_Pow(LL x, int n) {
    LL res = 1;
    while(n) {
        if(n & 1) res = res * x % mod;
        x = x * x % mod;
        n >>= 1;
    }
    return res;
}

void init() {
    A[0] = 1;
    for(int i = 1; i < maxn; i++) A[i] = (A[i-1] * i) % mod;
    inv[maxn-1] = Mod_Pow(A[maxn-1], mod - 2);
    for(int i = maxn - 2; i >= 0; i--) {
        inv[i] = inv[i+1] * (i + 1) % mod;
    }
}

int Com(int n, int m) {
    if(n < 0 || m < 0 || m > n) return 0;
    if(m == 0 || m == n) return 1;
    return ((A[n] * inv[n-m] % mod) * inv[m]) % mod;
}

int main() {
#ifndef ONLINE_JUDGE
    FIN;
#endif
    init();
    int icase = 0;
    scanf("%d", &t);
    while(t--) {
        scanf("%d%d", &n, &m);
        LL sum = 0;
        for(int i = 0; i <= m; i++) {
            if(i % 2 == 0) {
                sum = (sum + ((Com(m, i) * Mod_Pow(2, i) % mod) * A[2 * n - i]) % mod) % mod;
            } else {
                sum = (sum - ((Com(m, i) * Mod_Pow(2, i) % mod) * A[2 * n - i]) % mod + mod) % mod;
            }
        }
        printf("Case #%d: %lld
", ++icase, sum % mod);
    }
    return 0;
}