HDU 1695(容斥+欧拉函数+素数分解)

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hdu 1695
题目大意：
求 xϵ(1,n), yϵ(1,m), gcd(x,y)=k的x,y的对数；
思路:
令n<m
xϵ(1,n), yϵ(1,n), num1gcd(x,y)=φ(nk);
xϵ(1,n), yϵ(n+1,m), 因式分解x, num2=nk−num(1∼nk中x各因子的倍数), 使用容斥解决;
ans=num1+num2;

#include <iostream>#include <cstdio>#define LL __int64#define N 100002using namespace std;LL euler[N];struct Num{    int prim[20];    int num;}c[N];void init_Euler()    //euler打表+素数分解{    for (int i = 0; i < N; i++)    {        c[i].num = 0;        euler[i] = 0;    }    euler[1] = 1;    for (int i = 2; i < N; i++)    {        if (!euler[i])        {            for (int j = i; j < N; j += i)            {                if (!euler[j])                {                    euler[j] = j;                }                euler[j] = euler[j] / i * (i - 1);                c[j].prim[c[j].num] = i;                c[j].num++;            }        }        euler[i] += euler[i-1];    }}LL get_ans(int pi, LL n)   //枚举版{    LL ans = 0;    for (int i = 1; i < (1 << c[pi].num); i++)    {        int tmp = 1, flag = 0;        for (int j = 0; j < c[pi].num; j++)        {            if (i & (1 << j))            {                flag++;                tmp *= c[pi].prim[j];            }        }        if (flag & 1)        {            ans += n / tmp;        }        else        {            ans -= n / tmp;        }    }    return ans;}LL dfs(int index, LL b, LL n)   //递归版{    LL ans = 0, t;    for (int i = index; i < c[n].num; i++)    {        t = b / c[n].prim[i];        ans += t - dfs(i + 1, t, n);    }    return ans;}int main(){    int T;    init_Euler();    while (~scanf("%d", &T))    {        for (int cas = 1; cas <= T; cas++)        {            int a, c;            LL b, d, k;            scanf("%d%I64d%d%I64d%I64d", &a, &b, &c, &d, &k);            if (k == 0)            {                printf("Case %d: 0\n", cas);                continue;            }            if (b > d)            {                swap(b, d);            }            LL tmp = b / k;            LL ans = euler[tmp];            if (b == d)            {                printf("Case %d: %I64d\n", cas, ans);                continue;            }            b /= k;            d /= k;            for (int i = b + 1; i <= d; i++)            {                // ans += b - get_ans(i, b);                ans += b - dfs(0, b, i);            }            printf("Case %d: %I64d\n", cas, ans);        }    }    return 0;}

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