From 7e632e62e0eae1d5f95e483b831d1b8bfc418b9f Mon Sep 17 00:00:00 2001
From: Kirito <1362050620@qq.com>
Date: Wed, 12 Oct 2016 11:31:08 +0800
Subject: [PATCH] Create 1840_x7536110.cpp

---
 QUSTOJ/1840_x7536110.cpp | 56 ++++++++++++++++++++++++++++++++++++++++
 1 file changed, 56 insertions(+)
 create mode 100644 QUSTOJ/1840_x7536110.cpp

diff --git a/QUSTOJ/1840_x7536110.cpp b/QUSTOJ/1840_x7536110.cpp
new file mode 100644
index 0000000..c0c8142
--- /dev/null
+++ b/QUSTOJ/1840_x7536110.cpp
@@ -0,0 +1,56 @@
+#include <bits/stdc++.h>
+typedef long long ll;
+using namespace std;
+const int P = 1e9+7;
+const int maxn = 1e6+10;
+ll f[maxn], r[maxn];
+ll C(int n,int m){ return n < m? 0 : f[n]*r[n-m]%P*r[m]%P;}
+ll lucas(int n,int m){
+    if(n < m) return 0;
+    if(!m||n == m) return 1;
+    return C(n%P, m%P)*lucas(n/P, m/P)%P;
+}
+void init(){
+    int i;
+    for(r[0] = r[1] = f[0] = f[1] = 1, i = 2; i < maxn; i++){
+        f[i] = f[i-1]*i%P, r[i] = -r[P%i]*(P/i)%P;
+        while(r[i] < 0) r[i] += P;
+    }
+    for(i = 2; i < maxn; i++)
+        r[i] = r[i]*r[i-1]%P;
+}
+ll solve(int n, int m, int k){//n个位置链选m个，间隔 >= k
+    return C(n-(m-1)*k, m);
+}
+ 
+int main(){
+    //freopen("in", "r", stdin);
+    //freopen("out", "w", stdout);
+    init();
+    int T; scanf("%d", &T);
+    ll n, m, k;
+    while(T--){
+        scanf("%lld%lld%lld", &n, &m, &k);
+        if(m == 1) printf("%lld\n", n);
+        else if((k+1)*m > n) puts("0");
+        else {
+            ll ans = ( solve(n-2*k-1, m-1, k)*k%P + solve(n-k, m, k))%P;
+            printf("%lld\n", ans);
+        }
+    }
+}
+/*
+1张圆桌有n个不同的座位，你需要留出m个座位给学生安排座位。为了防止作弊，任意两个相邻学生坐的座位至少需要隔开k个空座位。
+问：有几种不同的选择座位的方式？方案数可能很大，需要对1e9+7取模。
+0 < m < n < 1e6, 0 < k < 1000, T <= 1000
+input
+2
+4 2 6
+5 2 1
+output
+0
+5
+第2组样例解释：
+座位编号为1,2,3,4,5。
+有2个宾客到来，中间隔1个座位。可以选择(1,3)(1,4)(2,4)(2,5)(3,5)这5种方式。
+*/