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

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#找规律 #素数筛法 #2015多校联合训练赛 #hdu

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CRB有N种不同的糖果，他想要吃K个糖果，想知道有多少种组合方式。这是一个数学问题，需要求出所有可能组合的最小公倍数。通过输入的测试案例数量和糖果种类数量，输出最小公倍数的模10^9+7的结果。

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

Sample Output

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 long
ll mod = 1000000007;
#define maxn 1000007
int 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;
}