LightOJ - 1067 Combinations
作者：互联网
题目描述：
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
3
4 2
5 0
6 4
Sample Output
Case 1: 6
Case 2: 1
Case 3: 15
题目大意：
从n个数当中取k个数，问有多少种取法？
解题报告：
1：先用dic数组将阶乘打表。
2：根据公式计算时，除法要转为乘上逆元。
代码：
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1000003;
const ll N = 1e6+10;
ll dic[N];
ll exgcd(ll a, ll b, ll& x, ll& y){
if(b == 0){
x = 1;
y = 0;
return a;
}
ll t = exgcd(b, a%b, y, x);
y -=a/b*x;
return t;
}
ll inv(ll a, ll mod){ // 求在mod下a的逆元
ll x, y;
exgcd(a, mod, x, y);
return (x%mod+mod)%mod;
}
void init(){
dic[0] = 1;
for(ll i=1; i<N; ++i){
dic[i] = dic[i-1]*i, dic[i] %= mod;
}
}
int main(){
ll t, cas = 1, n, k;
scanf("%lld", &t);
init();
while(t--){
scanf("%lld%lld", &n, &k);
printf("Case %lld: %lld\n", cas++, (dic[n]*inv(dic[k], mod)%mod)*inv(dic[n-k], mod)%mod );
}
return 0;
}
baronLJ
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标签：return,ll,LightOJ,exgcd,1067,dic,Take,Combinations,mod 来源： https://blog.csdn.net/jun_____/article/details/104070272