CF：358C 暴力DP筛选素数预处理 - c++编程基础

C. Bear and Prime Numbers time limit per test 2 seconds memory limit per test 512 megabytes input standard input output standard output

Recently, the bear started studying data structures and faced the following problem.

You are given a sequence of integers x1, x2, ..., xn of length n and m queries, each of them is characterized by two integers li, ri. Let's introduce f(p) to represent the number of such indexes k, that xk is divisible by p. The answer to the query li, ri is the sum: $\$ , where S(li, ri) is a set of prime numbers from segment [li, ri] (both borders are included in the segment).< http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">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"brush:java;">6 5 5 7 10 14 15 3 2 11 3 12 4 4 output

9
7
0

input

7
2 3 5 7 11 4 8
2
8 10
2 123

output

0
7

Note

Consider the first sample. Overall, the first sample has 3 queries.

The first query l = 2, r = 11 comes. You need to count f(2) + f(3) + f(5) + f(7) + f(11) = 2 + 1 + 4 + 2 + 0 = 9.
The second query comes l = 3, r = 12. You need to count f(3) + f(5) + f(7) + f(11) = 1 + 4 + 2 + 0 = 7.

The third query comes l = 4, r = 4. As this interval has no prime numbers, then the sum equals 0.

这题太神了！比赛的时候不知道怎么做，唉……没想到可以这样暴力的。太神了……递推的DP，查询时间复杂为O(1)，比线段树还快，这个能这样暴力真是服了，见识短浅啊……

dp[i]表示当前2到 i 这中间能被那n 个数整除的数之和……然后求 l 到 r 的时候就可以直接dp [ r ] - dp [ l-1] 了。

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                   #define PI acos(-1.0) #define mem(a,b) memset(a,b,sizeof(a)) #define sca(a) scanf("%d",&a) #define pri(a) printf("%d\n",a) #define lson i<<1,l,mid #define rson i<<1|1,mid+1,r #define MM 10000005 #define MN 3005 #define INF 10000007 #define eps 1e-7 using namespace std; typedef long long ll; int vis[MM],dp[MM],is[MM]; void getprime() { int i,j; for(i=2;i<=MM;i++) if(!is[i]) { if(vis[i]) dp[i]+=vis[i]; for(j=i+i;j<=MM;j+=i) { if(vis[j]) dp[i]+=vis[j]; is[j]=1; } } for(i=2;i<=MM;i++) dp[i]+=dp[i-1]; } int main() { int n,m,i,a,l,r; sca(n); for(i=0;i
                  
                   MM MM:l; r=r>MM MM:r; pri(dp[r]-dp[l-1]); } return 0; }