hdu4135

题意：

给出l,r,n，询问[l,r]与n互质的数的个数
询问共T组，保证0<T≤100，1≤n≤109,1≤l,r≤1015

solution：

用r的前缀和减去l−1的前缀和，将n暴力质因数分解
枚举n的质因子集合，每次统计有多少数字是子集乘积的倍数
这个东西用容斥原理直接算一下就好了

#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>#include<cstring>#include<vector>#include<queue>#include<set>#include<map>#include<stack>#include<bitset>#include<ext/pb_ds/priority_queue.hpp>using namespace std;const int maxn = 32000;typedef long long LL;int T,n,tot,tp,pri[maxn],stk[maxn];bool not_pri[maxn]; LL l,r;LL Calc(LL k){    LL ret = k;    for (int o = 1; o < (1 << tp); o++)    {        LL sum,tmp; sum = tmp = 1;        for (int j = 0; j < tp; j++)            if (o & (1 << j)) sum *= stk[j],tmp *= -1;        ret += tmp * (k / sum);    }    return ret;}void Solve(int I){    scanf("%lld%lld%d",&l,&r,&n);    for (int i = 1; i <= tot; i++)    {        if (n % pri[i] != 0) continue;        stk[tp++] = pri[i];        while (n % pri[i] == 0) n /= pri[i];        if (n == 1) break;    }    if (n > 1) stk[tp++] = n;    LL Ans = Calc(r);    if (l > 1) Ans -= Calc(l - 1);    printf("Case #%d: %lld\n",I,Ans); tp = 0;}int main(){    #ifdef DMC        freopen("DMC.txt","r",stdin);    #endif    for (int i = 2; i < maxn; i++)    {        if (!not_pri[i]) pri[++tot] = i;        for (int j = 1; j <= tot; j++)        {            int Nex = pri[j] * i;            if (Nex >= maxn) break;            not_pri[Nex] = 1;            if (i % pri[j] == 0) break;        }    }    cin >> T;    for (int i = 1; i <= T; i++) Solve(i);    return 0;}