lightoj 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=4and 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

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PROBLEM SETTER: JANE ALAM JAN

http://www2.chinaedu.com/101resource004/wenjianku/200501/101ktb/lanmu/XF1S0213/XF1S0213.htm

隔板法

思路：组合数学，费马小定理求逆元，快速幂。

用隔板法来求，这个问题可以转化为x1+x2+...xk=n的多元方不同程解的个数，并且xk〉=0；

就是组合数C(n+k-1,k-1) ，那么由费马小定理a^p-1==1mod(p);设a^-1为a的逆元则（a*a^-1*a^p-2）=a^-1mod（p）；

即a^p-2=a^-1mod（p）；C(a,b) =（f（a））/(f(b)*f(a-b));

C(a,b)%p=(（f（a））/(f(b)*f(a-b)))%P;其中f(n)表示阶乘。

（a/b）%p=k%p；两边同乘b ----a%p=(k*b)%p;然后两边同乘b^-1%p；----a*b^-1%p=k%p;

而根据费马小定理a^p-2=a^-1mod（p）；用快速幂求下b^p-2%p就可以了。

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<math.h>
#include<queue>
using namespace std;
const long long N=1e9+7;
typedef long long LL;
LL quickmi(long long  a,long long  b);
LL DP[2*1000005];
void Init()
{
    int i,j;
    DP[0]=1;
    for(i=1;i<=2000005;i++)
    {
        DP[i]=(DP[i-1]*i)%N;
    }
}
int main(void)
{Init();
    int i,j,k,p,q;
    scanf("%d",&k);
    for(i=1;i<=k;i++)
    {
        scanf("%d %d",&p,&q);
        LL ans=quickmi(DP[q-1]*DP[p]%N,N-2);
        printf("Case %d: ",i);
        printf("%lld
",DP[p+q-1]*ans%N);
    }
    return 0;
}

LL quickmi(long long  a,long long  b)
{
    LL sum=1;
    while(b)
    {
        if(b&1)
       sum=(sum*a)%(N);
       a=(a*a)%N;
       b/=2;
    }
    return sum;
}