Light OJ 1058

题意: 简单的就组合数 C(m,n);

数据多,大, 要预处理;

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 1e6 + 2;
const LL MOD = 1000003; LL Pow_Mod(LL a, LL n)
{
LL ret = 1;
while(n)
{
if(n & 1) ret = ret * a % MOD;
n >>=1;
a = a * a % MOD;
}
return ret;
} LL Num[maxn], Inv[maxn];
LL Init()
{
Num[0] = 1;
for(LL i = 1; i < maxn; ++i) Num[i] = Num[i-1] * i % MOD;
//cout << Num[maxn-1] << endl;
Inv[maxn-1] = Pow_Mod(Num[maxn-1], MOD-2);
for(LL i = maxn-2; i >= 0; --i)
Inv[i] = Inv[i+1] * (i+1) % MOD;
} LL C(LL m, LL n)
{
LL ret = 1;
if(n == 0 || n == m) return ret;
else
{
LL s = m-n;
//cout << s << endl;
ret = Num[m] * Inv[s] % MOD;
//cout << Num[m] << " " << Inv[s] << endl;
ret = ret * Inv[n] % MOD;
//cout << Inv[n] << endl;
return ret;
}
} int main()
{
Init();
int t;
LL m, n;
scanf("%d",&t);
for(int kase = 1; kase <= t; ++kase)
{
scanf("%lld %lld",&m,&n) ;
printf("Case %d: %lld\n",kase, C(m,n));
}
}
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