lightoj 1102 - Problem Makes Problem（组合数+逆元）

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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 3

3 5

1000 3

1000 5

Case 1: 15

Case 2: 35

Case 3: 501501

Case 4: 84793457

题意是给你一个数n，让你把它分成非负的k份。

就是方程n=x1+x2+x3+...+xk非负解的个数。可以看成把n+k分成k份，然后分出来的k个每个再减1，这样就可以用隔板法用k-1个板隔成k份，就是C（n+k-1，k-1）。

接下来主要就是求组合数了。C（a，b）% p = f[a]/(f[a-b]*f[b]) % p（f[]是阶乘）

因为除法不能取模，所以要利用乘法逆元。

这里可以直接求乘法逆元（用拓展欧几里得）

也可以用费马小定理：a^p-1==1mod(p)，所以逆元a^-1==a^p-2mod（p）,然后用快速幂求出来就可以了。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define LL long long#define mod 1000000007using namespace std;LL f[2000010];void init(){    f[0] = 1;    for(int i=1;i<=2000000;i++)        f[i] = f[i-1]*i%mod;}LL quickpow(LL x,int n){    LL res = 1;    while(n > 0)    {        if(n&1)            res = res*x%mod;        x = x*x%mod;        n >>= 1;    }    return res;}int main(void){    int T,n,k;    init();    scanf("%d",&T);    int cas = 1;    while(T--)    {        scanf("%d%d",&n,&k);        LL ans = quickpow(f[n]*f[k-1]%mod,mod-2);        ans = ans*f[n+k-1]%mod;        printf("Case %d: %lld\n",cas++,ans);    }    return 0;}

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