题目链接:http://www.lightoj.com/volume_showproblem.php?problem=1067
Given n different objects, you want to take k of them. How many ways to can do it?
For example, say there are 4 items; you want to take 2 of them. So, you can do it 6 ways.
Take 1, 2
Take 1, 3
Take 1, 4
Take 2, 3
Take 2, 4
Take 3, 4
Input
Input starts with an integer T (≤ 2000), denoting the number of test cases.
Each test case contains two integers n (1 ≤ n ≤ 106), k (0 ≤ k ≤ n).
Output
For each case, output the case number and the desired value. Since the result can be very large, you have to print the result modulo 1000003.
Sample Input |
Output for Sample Input |
3 4 2 5 0 6 4 |
Case 1: 6 Case 2: 1 Case 3: 15 |
分析:
时间只有2秒,T组测试数据加上n的106达到了109递推肯定超时,那么考虑组合公式,C(n,k)=n!/(k!*(n-k)!);先打一个阶乘的表(当然要取模,只有106),然后就是这个除法取模的问题,当然是求逆元,(a/b)%mod=a*(b对mod 的逆元);求逆元可以用扩欧和费马小定理。
费马小定理的使用条件mod必须为素数。
代码:
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
using namespace std;
#define N 1000009
#define mod 1000003
#define LL long long
LL fact[N];
void init()
{
fact[0] = fact[1] = 1;
for(int i = 2; i < N; i++)
fact[i] = (fact[i-1] * i) % mod;
}
/*LL niyuan(int a, int p)
{
LL ans = 1;
if(p == 0)
return 1;
while(p)
{
if(p & 1)
ans = (ans * a) % mod;
a = (a * a) % mod;
p >>= 1;
}
return ans;
}*/
LL niyuan(int a, int b)///就是一个快速幂。
{
if(b == 0)
return 1;
LL x = niyuan(a, b / 2);
LL ans = x * x % mod;
if(b % 2 == 1)
ans = ans * a % mod;
return ans;
}
LL c(int n, int k)
{
LL fm = (fact[k] * fact[n-k]) % mod;///n! * (n-m)!.
LL ans1 = niyuan(fm, mod - 2);///求n!的逆元。
return (ans1 * fact[n]) % mod;///公式(a / b ) % mod = (a * a ^(mod-2) % mod。
}
int main(void)
{
int T, cas;
int n, k;
init();
scanf("%d", &T);
cas = 0;
while(T--)
{
cas++;
scanf("%d%d", &n, &k);
if(2 * k > n)///组合数性质。
k = n - k;
printf("Case %d: %lld ", cas, c(n, k));
}
return 0;
}