hdu 6169 Senior PanⅡ Miller_Rabin素数测试+容斥-优快云博客

本文介绍了一道关于求解特定区间内整数最小除数之和的问题，并提供了详细的算法实现过程，包括质数判断、状态转移等关键步骤。

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Senior PanⅡ

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)

Problem Description

Senior Pan had just failed in his math exam, and he can only prepare to make up for it. So he began a daily task with Master Dong, Dong will give a simple math problem to poor Pan everyday.
But it is still sometimes too hard for Senior Pan, so he has to ask you for help.
Dong will give Pan three integers L,R,K every time, consider all the positive integers in the interval [L,R], you’re required to calculate the sum of such integers in the interval that their smallest divisor (other than 1) is K.

Input

The first line contains one integer T, represents the number of Test Cases.
Then T lines, each contains three integers L,R,K(1≤L≤R≤10^11,2≤K≤10^11)

Output

For every Test Case, output one integer: the answer mod 10^9+7

Sample Input

2 1 20 5 2 6 3

Sample Output

Case #1: 5 Case #2: 3

Source

2017 Multi-University Training Contest - Team 9

占坑；。。。其实是不想写，容斥我用莫比乌斯函数推出来的

突然发现我sb，为什么写大素数测试啊。。。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<bitset>
#include<set>
#include<map>
using namespace std;
#define LL long long
#define pi (4*atan(1.0))
#define eps 1e-8
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=3e3+10,M=1e6+10,inf=1e9+10;
const LL INF=1e18+10,mod=1e9+7;


LL gcd(LL a, LL b)
{
    return b? gcd(b, a % b) : a;
}

LL multi(LL a, LL b, LL m)
{
    LL ans = 0;
    a %= m;
    while(b)
    {
        if(b & 1)
        {
            ans = (ans + a) % m;
            b--;
        }
        b >>= 1;
        a = (a + a) % m;
    }
    return ans;
}

LL quick_mod(LL a, LL b, LL m)
{
    LL ans = 1;
    a %= m;
    while(b)
    {
        if(b & 1)
        {
            ans = multi(ans, a, m);
            b--;
        }
        b >>= 1;
        a = multi(a, a, m);
    }
    return ans;
}
const int Times = 10;
bool Miller_Rabin(LL n)
{
    if(n == 2) return true;
    if(n < 2 || !(n & 1)) return false;
    LL m = n - 1;
    int k = 0;
    while((m & 1) == 0)
    {
        k++;
        m >>= 1;
    }
    for(int i=0; i<Times; i++)
    {
        LL a = rand() % (n - 1) + 1;
        LL x = quick_mod(a, m, n);
        LL y = 0;
        for(int j=0; j<k; j++)
        {
            y = multi(x, x, n);
            if(y == 1 && x != 1 && x != n - 1) return false;
            x = y;
        }
        if(y != 1) return false;
    }
    return true;
}

int vis[M];
vector<int>pri;
void init()
{
    for(int i=2;i<=350000;i++)
    {
        if(!vis[i])
        pri.push_back(i);
        for(int j=i+i;j<=350000;j+=i)
            vis[j]=1;
    }
}
LL out;
void dfs(LL p,int pos,int step,LL L,LL R,LL K)
{
    if(p*K>R)return;
    LL d=p*K;
    LL x1=(R/d),x2=(L-1)/d;
    //cout<<p<<" "<<d<<" "<<step<<endl;
    //cout<<p<<" "<<pos<<" "<<step<<endl;
    if(step%2==0)
    {
        out+=((multi(x1,x1+1,mod)*1LL*500000004)%mod)*d;
        out%=mod;
        out-=((multi(x2,x2+1,mod)*1LL*500000004)%mod)*d;
        out=(out%mod+mod)%mod;
    }
    else
    {
        out-=((multi(x1,x1+1,mod)*1LL*500000004)%mod)*d;
        out=(out%mod+mod)%mod;
        out+=((multi(x2,x2+1,mod)*1LL*500000004)%mod)*d;
        out%=mod;
    }
    for(int i=pos;i<pri.size();i++)
    {
        if(pri[i]>=K)return;
        if(p*pri[i]*K>R)return;
        dfs(p*pri[i],i+1,step+1,L,R,K);
    }
}
int main()
{
    init();
    int T,cas=1;
    scanf("%d",&T);
    while(T--)
    {
        out=0;
        LL l,r,k;
        scanf("%lld%lld%lld",&l,&r,&k);
        bool isp=Miller_Rabin(k);
        printf("Case #%d: ",cas++);
        if(!isp){
            printf("0\n");
            continue;
        }
        if(k>=350000)
        {
            if(l<=k&&r>=k)printf("%lld\n",k%mod);
            else printf("%lld\n",0);
        }
        else
        {
            dfs(1,0,0,l,r,k);
            printf("%lld\n",out);
        }
    }
    return 0;
}