LOJ#2095 选数

给定n，k，l，r

问从[l, r]中选出n个数gcd为k的方案数。

解：稍微一想就能想到反演，F(x)就是[l, r]中x的倍数个数的n次方。

后面那个莫比乌斯函数随便怎么搞都行，当然因为这是杜教筛的题就杜教筛了。

然后写一写，交上去80分......

然后枚举一下d是k的多少倍，我们发现F(x)的计算式[(r / x) - ((l - 1) / x)]ⁿ可以整除分块...

然后就做完了。

 #include <cstdio>
 #include <map>
 #include <algorithm>

 typedef long long LL;
 const int N = , T = ;
 const LL MO = ;

 std::map<LL, LL> mp;
 int p[N], top, miu[N];
 LL n, k, l, r, Miu[N], L, R;
 bool vis[N];

 inline void getp(int n) {
     miu[] = ;
     for(int i = ; i <= n; i++) {
         if(!vis[i]) {
             p[++top] = i;
             miu[i] = -;
         }
         for(int j = ; j <= top && i * p[j] <= n; j++) {
             vis[i * p[j]] = ;
             if(i % p[j] == ) {
                 break;
             }
             miu[i * p[j]] = -miu[i];
         }
     }
     for(int i = ; i <= n; i++) {
         Miu[i] = Miu[i - ] + miu[i];
     }
     return;
 }

 inline LL qpow(LL a, LL b) {
     LL ans = ;
     a %= MO;
     while(b) {
         if(b & ) ans = ans * a % MO;
         a = a * a % MO;
         b = b >> ;
     }
     return ans;
 }

 LL getMiu(LL x) {
     if(x <= ) return ;
     if(x <= T) return Miu[x];
     if(mp.count(x)) return mp[x];
     LL ans = ;
     for(LL i = , j; i <= x; i = j + ) {
         j = x / (x / i);
         ans -= (j - i + ) % MO * getMiu(x / i) % MO;
         ans %= MO;
     }
     return mp[x] = (ans + MO) % MO;
 }

 LL F(LL x) {
     LL temp = (R / x) - (L / x);
     return qpow(temp, n);
 }

 int main() {
     getp(T);
     scanf("%lld%lld%lld%lld", &n, &k, &l, &r);
     LL ans = ;
     L = (l - ) / k, R = r / k;
     for(LL i = , j; i <= R; i = j + ) {
         if(L / i) j = std::min(R / (R / i), L / (L / i));
         else if(R / i) j = R / (R / i);
         else j = R;
         ans += F(i) * (getMiu(j) - getMiu(i - )) % MO;
         ans %= MO;
     }
     printf("%lld\n", (ans + MO) % MO);

     return ;
 }

AC代码

整除分块的时候要判断是不是0啊......