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题目大意：给你一个n(10^8)以内，让你求出1-n中与n互质的数x^4的和。

解题思路：先把n进行分解质因数，然后容斥求出所有与n不互质的数x^4的和，然后做减法用总的减去不互质的就是互质的。

注意：1^4+2^4+……+n^4 = n(n+1)(2n+1)(3n^2+3n-1)/30.

The Boss on Mars

2
4
5

82
354
Hint
Case1: sum=1+3*3*3*3=82
Case2: sum=1+2*2*2*2+3*3*3*3+4*4*4*4=354
 

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-8
#define M 1000100
#define LL long long
//#define LL long long
#define INF 0x3f3f3f
#define PI 3.1415926535898
#define mod 1000000007


const int maxn = 110;

using namespace std;
///bool f[maxn];

int num[maxn];
int ans;
void Find(LL n)
{
    ans = 0;
    for(int i = 2; i*i <= n; i++)
    {
        if(n%i==0)
        {
            num[ans++] = i;
            while(n%i == 0) n /= i;
        }
    }
    if(n > 1) num[ans++] = n;
}

LL fastmod(LL a, LL k)
{
    LL b = 1LL;
    while(k)
    {
        if(k&1) b = a*b%mod;
        a = (a%mod)*(a%mod)%mod;
        k /= 2;
    }
    return b;
}

LL Pow(LL n, int k)
{
    LL ff = 1LL;
    for(int i = 0; i < k; i++) 
    {
        ff *= n;
        if(ff > mod) ff %= mod;
    }
    return ff%mod;
}

LL Get(LL n)
{
    LL sum = n*(n+1)%mod;
    sum %= mod;
    sum *= (2*n+1)%mod;
    sum %= mod;
    sum *= ((3*(n*n%mod))%mod+((3*n)%mod)-1)%mod;
    sum %= mod;
    LL sp = fastmod(30, mod-2);
    sum *= sp;
    sum %= mod;
    return sum;
}

int main()
{
    int T;
    cin >>T;
    while(T--)
    {
        LL n;
        scanf("%I64d",&n);
        Find(n);
        LL sum = 0;
        LL xsum;
        LL xans = Get(n);
        for(int i = 1; i < (1<<ans); i++)
        {
            LL sx = 1;
            int s = 0;
            for(int j = 0; j < ans; j++)
            {
                if(i&(1<<j))
                {
                    sx *= num[j];
                    s ++;
                }
            }
            if(s%2)
            {
                xsum = Pow(sx, 4);
                xsum %= mod;
                sum += (xsum*Get(n/sx))%mod;
                sum %= mod;
            }
            else
            {
                xsum = Pow(sx, 4);
                xsum %= mod;
                sum -= (xsum*Get(n/sx))%mod;
                sum %= mod;
                while(sum < 0) sum += mod;
            }
        }
        if(n == 1)
        {
            cout<<0<<endl;
            continue;
        }
        xans -= sum;
        while(xans < 0) xans += mod;
        printf("%I64d\n",xans%mod);
    }
}

HDU 4059 The Boss on Mars(数论)

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原文地址：http://blog.csdn.net/xu12110501127/article/details/41344333