hdu 5407 CRB and Candies 2015多校联合训练赛#10 找规律 素数筛法

CRB and Candies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 402    Accepted Submission(s): 185

Problem Description

CRB has N different candies. He is going to eat K candies.
He wonders how many combinations he can select.
Can you answer his question for all K(0 ≤ K ≤ N)?
CRB is too hungry to check all of your answers one by one, so he only asks least common multiple(LCM) of all answers.

 

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case there is one line containing a single integer N.
1 ≤ T ≤ 300
1 ≤ N ≤ 106

 

Output

For each test case, output a single integer – LCM modulo 1000000007(109+7).

 

Sample Input

512345

 

Sample Output

1231210

 

Author

KUT（DPRK）

 

Source

2015 Multi-University Training Contest 10

 

居然可以找到规律呢！！我们算是懵了，没想过找规律的。哭哭哭~~~

#include<iostream>#include<cstring>#include<algorithm>#include<cstdio>using namespace std;#define ll long longll mod = 1000000007;#define maxn 1000007int check[maxn];ll LCM[maxn],ni[maxn];ll cal(ll a,int b){    ll ans = 1;    while(b){        if(b&1) ans = ans*a%mod;        b /= 2;        a = a*a%mod;    }    return ans;}void init(){    memset(check,0,sizeof(check));    for(int i = 2;i < maxn; i++){        if(check[i] == 0){            for(int j = i;j < maxn; j+=i)                check[j] = 1;            for(ll j = i;j < maxn; j = j*i)                check[j] = i;        }    }    LCM[1] = 1;    for(int i = 2;i < maxn; i++){        LCM[i] = LCM[i-1]*check[i]%mod;    }    for(int i = 1; i < maxn; i++)        ni[i] = cal(i,mod-2);}int main(){    init();    int t,n;    scanf("%d",&t);    while(t--){        scanf("%d",&n);        ll ans = LCM[n+1]*ni[n+1]%mod;        printf("%I64d\n",ans);    }    return 0;}