LightOJ 1067 - Combinations (Lucas定理)
2016-02-23 00:07
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1067 - Combinations
Given n different objects, you want to take kof them. How many ways to can do it?
For example, say there are 4 items; you want to take 2 ofthem. 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
Each test case contains two integers n (1 ≤ n≤ 106), k (0 ≤ k ≤ n).
1000003.
题意:看样例就能看出题意了,C(n,m)%1000003
思路:1000003是素数,符合Lucas定理
ac代码:
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
LL gcd(LL a,LL b){return b?gcd(b,a%b):a;}
LL lcm(LL a,LL b){return a/gcd(a,b)*b;}
LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
//head
LL f[MAXN];
void findinit(LL p)
{
f[0]=1;
for(int i=1;i<=p;i++)
f[i]=(f[i-1]*i)%p;
}
LL lucas(LL n,LL m,LL p)
{
LL ans=1;
while(n&&m)
{
LL a=n%p,b=m%p;
if(a<b)
return 0;
ans=(ans*f[a]*powmod(f[b]*f[a-b]%p,p-2,p))%p;
n/=p;m/=p;
}
return ans;
}
int main()
{
findinit(1000003);
LL n,m;
int t,cas=0;
scanf("%d",&t);
while(t--)
{
scanf("%lld%lld",&n,&m);
LL ans=lucas(n,m,1000003);
printf("Case %d: ",++cas);
printf("%lld\n",ans);
}
return 0;
}
![]() ![]() | PDF (English) | Statistics | Forum |
Time Limit: 2 second(s) | Memory Limit: 32 MB |
For example, say there are 4 items; you want to take 2 ofthem. 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 modulo1000003.
Sample Input | Output for Sample Input |
3 4 2 5 0 6 4 | Case 1: 6 Case 2: 1 Case 3: 15 |
题意:看样例就能看出题意了,C(n,m)%1000003
思路:1000003是素数,符合Lucas定理
ac代码:
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
LL gcd(LL a,LL b){return b?gcd(b,a%b):a;}
LL lcm(LL a,LL b){return a/gcd(a,b)*b;}
LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
//head
LL f[MAXN];
void findinit(LL p)
{
f[0]=1;
for(int i=1;i<=p;i++)
f[i]=(f[i-1]*i)%p;
}
LL lucas(LL n,LL m,LL p)
{
LL ans=1;
while(n&&m)
{
LL a=n%p,b=m%p;
if(a<b)
return 0;
ans=(ans*f[a]*powmod(f[b]*f[a-b]%p,p-2,p))%p;
n/=p;m/=p;
}
return ans;
}
int main()
{
findinit(1000003);
LL n,m;
int t,cas=0;
scanf("%d",&t);
while(t--)
{
scanf("%lld%lld",&n,&m);
LL ans=lucas(n,m,1000003);
printf("Case %d: ",++cas);
printf("%lld\n",ans);
}
return 0;
}
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