Codeforces Round #226 (Div. 2) C. Bear and Prime Numbers(暴力)
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题目链接:点击打开链接
题意:给n个数, m次询问, 每次询问时一个区间[l, r]。 求这个区间中的所有素数,能被n个数中的几个数整除的累加和。
思路:没想到什么好办法, 直接筛出素数以后直接暴力的,没想到跑的这么快。。
最费时间的大概就是那个二重循环, 但是可能因为素数比较少, 所以实际的时间复杂度并不高。
细节参见代码:
#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#include<cstdlib>#include<cmath>#include<set>#include<list>#include<deque>#include<map>#include<queue>#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))using namespace std;typedef long long ll;const double PI = acos(-1.0);const double eps = 1e-6;const int mod = 1000000000 + 7;const int INF = 1000000000;const int maxn = 10000000 + 10;int T,n,m,maxd,c,a[(int)1e6+10],vis[maxn],cnt[maxn],hehe[maxn],ans[(int)1e6+10];void init() { int m = sqrt(maxd+0.5); c = 0; for(int i=2;i<=m;i++) if(!vis[i]) { for(int j=i*i;j<=maxd;j+=i) vis[j] = true; } for(int i=2;i<=maxd;i++) { if(!vis[i]) { for(int j=i;j<=maxd;j+=i) { cnt[i] += hehe[j]; } } }}int main() { scanf("%d",&n); maxd = 0; for(int i=0;i<n;i++) { scanf("%d",&a[i]); maxd = max(maxd, a[i]); hehe[a[i]]++; } init(); for(int i=2;i<=maxd;i++) { ans[i] = ans[i-1] + cnt[i]; } scanf("%d",&m); int l, r; while(m--) { scanf("%d%d",&l,&r); if(r > maxd) r = maxd; if(l > maxd) printf("0\n"); else printf("%d\n",ans[r]-ans[l-1]); } return 0;}
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