zoukankan      html  css  js  c++  java
  • HDU 5407——CRB and Candies——————【逆元+是素数次方的数+公式】

    CRB and Candies

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 722    Accepted Submission(s): 361


    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
    5
    1 2 3 4 5
     
    Sample Output
    1
    2
    3
    12
    10
     
    Author
    KUT(DPRK)
     
    Source

    题目大意:让你求LCM(C(n,0),C(n,1),C(n,2)...C(n,n-1),C(n,n)),最后结果取模。

    解题思路:其实只要有公式了,问题就很好解决了。f(n)是求1 - n的最小公倍数。这个是可以借鉴得。如果n是一个素数p的k次方,那么就乘以素数p。主要需要求逆元,和快速判断x是否为素数p的k次方。

    #include<bits/stdc++.h>
    using namespace std;
    typedef long long INT;
    const int maxn=1e6+20;
    const INT MOD=1e9+7;
    INT f[maxn],g[maxn],inv[maxn];
    int p[maxn];
    void init(){
        for(int i=1;i<maxn;i++){
            p[i]=i;
        }
        for(int i=2;i<maxn;i++){
            if(p[i]==i){
                for(int j=i+i;j<maxn;j+=i){
                    p[j]=i;
                }
            }
        }
    }
    bool check(int x){
        int d=p[x];
        if(x>1){
            while(x%d==0){
                x/=d;
            }
            return x==1;
        }
        return false;
    }
    void get_f(){
        f[1]=1;
        for(int i=2;i<maxn;i++){
            if(check(i)){
                f[i]=f[i-1]*p[i]%MOD;
            }else{
                f[i]=f[i-1];
            }
        }
    }
    INT Powmod(INT a,INT n){
        a%=MOD;
        INT ret=1;
        while(n){
            if(n&1)
                ret= ret * a % MOD;
            n>>=1;
            a = (a*a)%MOD;
        }
        return ret;
    }
    INT get_inv(int n){
        return Powmod((INT)n,MOD-2);
    }
    INT get_g(int n){
        return f[n+1]*get_inv(n+1)%MOD;
    }
    int main(){
        int t,n;
        init();
        get_f();
        scanf("%d",&t);
        while(t--){
            scanf("%d",&n);
            INT ans=get_g(n);
            printf("%lld
    ",ans);
        }
        return 0;
    }
    

      

  • 相关阅读:
    把本地的jar包安装到maven库中
    mybatis 查询
    freemarker
    python——线程相关
    【python-sql】sql操作时遇到的坑
    专项测试——移动app安装包检测
    【Android端 adb相关】adb相关总结
    for 语句
    Python2.x与3​​.x版本区别
    Python 运算符
  • 原文地址:https://www.cnblogs.com/chengsheng/p/4803332.html
Copyright © 2011-2022 走看看