标签：acm   数学   

题目：

CRB and Candies

5
1
2
3
4
5

1
2
3
12
10

题意：求出LCM{C(N,0),C(N,1),....,C(N,N)}%1000000007。

思路：打表找出前几项，然后在OEIS上发现LCM{C(N,0)...C(N,N)}=LCM(1,2,3,...,N)/N,所以我们只要求出LCM{1,2,3...,N},然后再除以n即可。算LCM的时候只要找出每个小于N的质数的次数最大值，然后把他们乘起来就可以了。除以n的时候由于要取模，所以变为乘以n对与mod的逆元。

代码：

#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include<climits>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>
using namespace std;

#define PB push_back
#define MP make_pair

#define REP(i,x,n) for(int i=x;i<(n);++i)
#define FOR(i,l,h) for(int i=(l);i<=(h);++i)
#define FORD(i,h,l) for(int i=(h);i>=(l);--i)
#define SZ(X) ((int)(X).size())
#define ALL(X) (X).begin(), (X).end()
#define RI(X) scanf("%d", &(X))
#define RII(X, Y) scanf("%d%d", &(X), &(Y))
#define RIII(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z))
#define DRI(X) int (X); scanf("%d", &X)
#define DRII(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define DRIII(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define OI(X) printf("%d",X);
#define RS(X) scanf("%s", (X))
#define MS0(X) memset((X), 0, sizeof((X)))
#define MS1(X) memset((X), -1, sizeof((X)))
#define LEN(X) strlen(X)
#define F first
#define S second
#define Swap(a, b) (a ^= b, b ^= a, a ^= b)
#define Dpoint  strcut node{int x,y}
#define cmpd int cmp(const int &a,const int &b){return a>b;}

 /*#ifdef HOME
    freopen("in.txt","r",stdin);
    #endif*/
const int MOD = 1e9+7;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<double> VD;
typedef long long LL;
typedef pair<int,int> PII;
//#define HOME

int Scan()
{
	int res = 0, ch, flag = 0;

	if((ch = getchar()) == '-')				//判断正负
		flag = 1;

	else if(ch >= '0' && ch <= '9')			//得到完整的数
		res = ch - '0';
	while((ch = getchar()) >= '0' && ch <= '9' )
		res = res * 10 + ch - '0';

	return flag ? -res : res;
}
/*----------------PLEASE-----DO-----NOT-----HACK-----ME--------------------*/



int vis[1000000+5];
int prime[1000000+5];
int cnt;
void getprime()
{
    cnt=0;
    MS0(vis);
    for(int i=2;i<=1000000;i++)
    {
     if(!vis[i])
            prime[cnt++]=i;
     for(int j=0;j<cnt&&prime[j]<=1000000/i;j++)
     {vis[prime[j]*i]=1;
     if(i%prime[j]==0)
        break;
    }
    }
}
long long inv(long long a,long long m)
{
    if(a==1)
        return 1;
    return inv(m%a,m)*(m-m/a)%m;
}
const int mod= 1e9+7;
int main()
{getprime();
int T;
RI(T);
while(T--)
{
    int n;
    RI(n);
    n++;
    int ans=1;
    for(int i=0;i<cnt&&prime[i]<=n;i++)
    {
        //int k=(int)(log10(n)/log10(prime[i])+0.5);
        int tmp=1;
        for(;(long long )tmp*prime[i]<=n;)
         tmp=((long long )tmp*prime[i])%mod;
        ans=((long long )ans*tmp)%mod;
    }
    ans=((ans*inv(n,mod))%mod+mod)%mod;
    printf("%d\n",ans);
}



        return 0;
}

hdu 5407 CRB and Candies

标签：acm   数学   

原文地址：http://blog.csdn.net/u013840081/article/details/48132107