标签:数字 first ret tmp oid 超过 its ons long
思路:先把所有幸运数字找出来, 把没有用的去掉,然后爆搜容斥,因为最多只会搜十几个就超过限制了,
所以是可行的。
#include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PII pair<int, int> #define y1 skldjfskldjg #define y2 skldfjsklejg using namespace std; const int N = 5000 + 7; const int M = 5e5 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1000000007; LL a[N], b[N], L, R, ans; int tot, top; void init(LL val) { if(val > R) return; if(val) a[++tot] = val; init(val * 10 + 6); init(val * 10 + 8); } void dfs(int p, int cnt, LL lcm) { if(p > top) { if(cnt & 1) ans += R / lcm - (L - 1) / lcm; else if(cnt) ans -= R / lcm - (L - 1) / lcm; return; } dfs(p + 1, cnt, lcm); long long tmp = lcm / __gcd(b[p], lcm); if(1.0 * b[p] * tmp <= R) dfs(p + 1, cnt + 1, b[p] * tmp); } int main() { scanf("%lld%lld", &L, &R); init(0); sort(a + 1, a + 1 + tot); for(int i = 1; i <= tot; i++) { bool flag = true; for(int j = 1; j <= top; j++) { if(a[i] % b[j] == 0) { flag = false; break; } } if(flag) b[++top] = a[i]; } reverse(b + 1, b + 1 + top); dfs(1, 0, 1); printf("%lld\n", ans); return 0; }
标签:数字 first ret tmp oid 超过 its ons long
原文地址:https://www.cnblogs.com/CJLHY/p/9593649.html