mirror of
https://github.com/Kiritow/OJ-Problems-Source.git
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57 lines
1.5 KiB
C++
57 lines
1.5 KiB
C++
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#include <bits/stdc++.h>
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typedef long long ll;
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using namespace std;
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const int P = 1e9+7;
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const int maxn = 1e6+10;
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ll f[maxn], r[maxn];
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ll C(int n,int m){ return n < m? 0 : f[n]*r[n-m]%P*r[m]%P;}
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ll lucas(int n,int m){
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if(n < m) return 0;
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if(!m||n == m) return 1;
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return C(n%P, m%P)*lucas(n/P, m/P)%P;
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}
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void init(){
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int i;
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for(r[0] = r[1] = f[0] = f[1] = 1, i = 2; i < maxn; i++){
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f[i] = f[i-1]*i%P, r[i] = -r[P%i]*(P/i)%P;
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while(r[i] < 0) r[i] += P;
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}
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for(i = 2; i < maxn; i++)
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r[i] = r[i]*r[i-1]%P;
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}
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ll solve(int n, int m, int k){//n个位置链选m个，间隔 >= k
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return C(n-(m-1)*k, m);
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}
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int main(){
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//freopen("in", "r", stdin);
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//freopen("out", "w", stdout);
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init();
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int T; scanf("%d", &T);
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ll n, m, k;
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while(T--){
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scanf("%lld%lld%lld", &n, &m, &k);
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if(m == 1) printf("%lld\n", n);
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else if((k+1)*m > n) puts("0");
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else {
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ll ans = ( solve(n-2*k-1, m-1, k)*k%P + solve(n-k, m, k))%P;
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printf("%lld\n", ans);
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}
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}
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}
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/*
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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。
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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。
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0 < m < n < 1e6, 0 < k < 1000, T <= 1000
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input
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2
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4 2 6
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5 2 1
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output
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0
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5
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第2组样例解释：
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座位编号为1,2,3,4,5。
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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。
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*/
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