zoukankan      html  css  js  c++  java
  • HDU 6390

    GuGuFishtion

    Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 1699    Accepted Submission(s): 627


    Problem Description
    Today XianYu is too busy with his homework, but the boring GuGu is still disturbing him!!!!!!
    At the break time, an evil idea arises in XianYu's mind.
    ‘Come on, you xxxxxxx little guy.’
    ‘I will give you a function ϕ(x) which counts the positive integers up to x that are relatively prime to x.’
    ‘And now I give you a fishtion, which named GuGu Fishtion, in memory of a great guy named XianYu and a disturbing and pitiful guy GuGu who will be cooked without solving my problem in 5 hours.’
    ‘The given fishtion is defined as follow:
    Gu(a,b)=ϕ(ab)ϕ(a)ϕ(b)

    And now you, the xxxxxxx little guy, have to solve the problem below given m,n,p.’
    (a=1mb=1nGu(a,b))(modp)

    So SMART and KINDHEARTED you are, so could you please help GuGu to solve this problem?
    ‘GU GU!’ GuGu thanks.
     
    Input
    Input contains an integer T indicating the number of cases, followed by T lines. Each line contains three integers m,n,p as described above.
    1T3
    1m,n1,000,000
    max(m,n)<p1,000,000,007
    And given p is a prime.
     
    Output
    Please output exactly T lines and each line contains only one integer representing the answer.
     
    Sample Input
    1
    5 7 23
     
    Sample Output
    2
    解析  我们容易得到 g(a,b)=ϕ(ab)/ϕ(a)ϕ(b)=gcd(a,b)/ϕ(gcd(a,b)
    1-n,1-m的 gcd最多有min(n,m)个所以转换成求1-n,1-m中gcd(a,b)=k,k的出现次数.
    AC代码
    #include <bits/stdc++.h>
    #define pb push_back
    #define mp make_pair
    #define fi first
    #define se second
    #define all(a) (a).begin(), (a).end()
    #define fillchar(a, x) memset(a, x, sizeof(a))
    #define huan printf("
    ")
    #define debug(a,b) cout<<a<<" "<<b<<" "<<endl
    #define ffread(a) fastIO::read(a)
    using namespace std;
    typedef long long ll;
    typedef pair<int,int> pii;
    const int maxn=1e6+10,inf=0x3f3f3f3f;
    int check[maxn],phi[maxn],prime[maxn];
    ll inv[maxn],f[maxn];
    void philist()
    {
        int cnt=0;
        phi[1]=1;
        for(int i=2; i<=maxn; i++)
        {
            if(!check[i])
            {
                prime[++cnt]=i;
                phi[i]=i-1;
            }
            for(int j=1; j<=cnt&&prime[j]*i<=maxn; j++)
            {
                check[i*prime[j]]=1;
                if(i%prime[j])
                    phi[i*prime[j]]=phi[i]*(prime[j]-1);
                else
                {
                    phi[i*prime[j]]=phi[i]*prime[j];
                    break;
                }
            }
        }
    }
    void init(int mod,int n)
    {
        fillchar(f,0);
        inv[1]=1;
        for(int i=2;i<=n;i++)
            inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
    }
    int main()
    {
        int t,n,m,p;
        scanf("%d",&t);
        philist();
        while(t--)
        {
            scanf("%d%d%d",&n,&m,&p);
            ll ans=0;
            if(n>m)
                swap(n,m);
            init(p,n);
            for(ll i=n;i>=1;i--)       //容斥求1-n,1-m中gcd为i的个数
            {
                f[i]=(ll)(n/i)*(m/i);
                for(int j=i+i; j<=n; j+=i) //f[i]表示gcd为i的对数有f[i]个
                    f[i]=f[i]-f[j];
                ans=(ans+f[i]%p*i%p*inv[phi[i]])%p;
            }
            printf("%lld
    ",ans);
        }
    }
  • 相关阅读:
    【转】JSP三种页面跳转方式
    我要从头做起
    转载:用 Tomcat 和 Eclipse 开发 Web 应用程序
    html的style属性
    Java连接oracle数据库
    tomcat遇到的问题(总结)
    ceshi
    今天要小结一下
    argument.callee.caller.arguments[0]与window.event
    JavaScript事件冒泡简介及应用
  • 原文地址:https://www.cnblogs.com/stranger-/p/9807329.html
Copyright © 2011-2022 走看看