询问区间$(l, r)$中有多少个数是只含$6, 8$的数的倍数
思路很妙啊。
首先在$10^{10}$内只含$6, 8$的数有$\sum_{i = 1}^{10} 2^i = 2046$个。
然后去掉相同的,应该是有$943$个。
之间算不好算,考虑用容斥原理。
但是直接容斥的复杂度很显然是$2^n$的
考虑剪枝!
当前的数大于上界,肯定要return
由于题目规定的数在$\sqrt {10^{10}}$内也就十几个,因此是可以跑过的。
注意中间算lcm的时候是会爆long long的!我们需要先转成double,再判断
// luogu-judger-enable-o2
// luogu-judger-enable-o2
// luogu-judger-enable-o2
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<stack>
#include<vector>
#include<cstring>
#define int long long
//#define int long long
using namespace std;
const int MAXN = 1e6;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9'){if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int Mx = 10000000000LL;
vector<int> a;
int flag[MAXN], vis[MAXN], b[MAXN], out, cnt;
void Pre(int now) {
if(now >= Mx) return ;
a.push_back(now);
Pre(now * 10 + 6); Pre(now * 10 + 8);
}
void dfs(int x, int opt, int l, int r, int pre) {
if(x > r) return ;
//printf("%I64d\n", out);
out += opt * (r / x - l / x);
for(int i = pre + 1; i <= cnt; i++) {
int g = __gcd(x, b[i]);
if((double) (x / g) * b[i] > (double)r) return;
dfs(x / g * b[i], opt * -1, l, r, i);
}
}
int solve(int l, int r) {
out = 0;
for(int i = 1; i <= cnt; i++)
dfs(b[i], 1, l, r, i);
return out;
}
main() {
Pre(6);
Pre(8);
//printf("%d", a.size());
for(int i = 0; i < a.size(); i++)
for(int j = 0; j < a.size(); j++)
if((i != j) && (a[j] > a[i]) && (a[j] % a[i] == 0)) flag[j] = 1;
//printf("%d\n", si);
for(int i = 0; i < a.size(); i++)
if(!flag[i]) b[++cnt] = a[i];
sort(b + 1, b + cnt + 1, greater<int>());
//for(int i = 1; i <= cnt; i++)
//- printf("%d\n", cnt);
int a = read(), b = read();
printf("%lld", solve(a - 1, b) + 1);
return 0;
}
/*
2 10000000000
*/