light oj 1102

1102 - Problem Makes Problem

As I am fond of making easier problems, I discovered a problem. Actually, the problem is 'how can you make n by adding k non-negative integers?' I think a small example will make things clear. Suppose n=4 and k=3. There are 15 solutions. They are

1.      0 0 4

2.      0 1 3

3.      0 2 2

4.      0 3 1

5.      0 4 0

6.      1 0 3

7.      1 1 2

8.      1 2 1

9.      1 3 0

10.  2 0 2

11.  2 1 1

12.  2 2 0

13.  3 0 1

14.  3 1 0

15.  4 0 0

As I have already told you that I use to make problems easier, so, you don't have to find the actual result. You should report the result modulo 1000,000,007.

Input

Input starts with an integer T (≤ 25000), denoting the number of test cases.

Each case contains two integer n (0 ≤ n ≤ 10⁶) and k (1 ≤ k ≤ 10⁶).

Output

For each case, print the case number and the result modulo 1000000007.

Sample Input	Output for Sample Input
4 4 3 3 5 1000 3 1000 5	Case 1: 15 Case 2: 35 Case 3: 501501 Case 4: 84793457

分析：

题目意思是把 n个元素分成k组且允许有空位置， 这就用到隔板法中的允许若干个人（或位置）为空的问题， 因为把元素分成k组需要k-1个隔板，并且可以允许元素个数为空，所以隔板可以放在任意位置，隔板加上元素个数一共有n+k-1个位置，那么就相当于从n+k-1个位置中选出k-1个位置放隔板即c（n-k+1， k-1）。然后直接用费小马定理（a/b)%mod = a * (b(^mod-2))%mod;求下逆元就可以了。

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#define N 2000010
#define mod 1000000007
using namespace std;
long long d[N];
void init()
{
d[0] = 1;

for(int i = 1; i < N; i++)
d[i] = (i * d[i-1]) % mod;
}

long long quickmi(long long a, long long b)
{
long long sum = 1;

while(b)
{
if(b & 1)
sum = (sum * a) % mod;
a = (a * a) % mod;
b /= 2;
}

return sum;
}
int main(void)
{
int T , cas;
int n, k;
scanf("%d", &T);
init();
cas = 0;

while(T--)
{
cas++;
scanf("%d%d", &n, &k);

long long ans = quickmi((d[k-1] * d[n]) % mod, mod-2);
ans = (d[n+k-1] * ans ) % mod;

printf("Case %d: %lld ", cas, ans);
}
return 0;
}