2018省赛赛第一次训练题解和ac代码

第一次就去拉了点思维很神奇的CF题目

2018省赛赛第一次训练

#	Origin	Title
		A	CodeForces 607A	Chain Reaction
		B	CodeForces 385C	Bear and Prime Numbers
		C	CodeForces 670D2	Magic Powder - 2
		D	CodeForces 360B	Levko and Array
		E	CodeForces 68B	Energy exchange
		F	CodeForces 24E	Berland collider
		G	CodeForces 429B	Working out
		H	CodeForces 329C	Graph Reconstruction
		I	CodeForces 329D	The Evil Temple and the Moving Rocks
		J	CodeForces 182E	Wooden Fence
		K	CodeForces 590D	Top Secret Task
		L	CodeForces 71E	Nuclear Fusion
		M	CodeForces 138D	World of Darkraft

A - Chain Reaction

 CodeForces - 607A 

There are n beacons located at distinct positions on a number line. The i-th beacon has position a_i and power level b_i. When the i-th beacon is activated, it destroys all beacons to its left (direction of decreasing coordinates) within distance b_i inclusive. The beacon itself is not destroyed however. Saitama will activate the beacons one at a time from right to left. If a beacon is destroyed, it cannot be activated.

Saitama wants Genos to add a beacon strictly to the right of all the existing beacons, with any position and any power level, such that the least possible number of beacons are destroyed. Note that Genos's placement of the beacon means it will be the first beacon activated. Help Genos by finding the minimum number of beacons that could be destroyed.

Input

The first line of input contains a single integer n (1 ≤ n ≤ 100 000) — the initial number of beacons.

The i-th of next n lines contains two integers a_i and b_i (0 ≤ a_i ≤ 1 000 000, 1 ≤ b_i ≤ 1 000 000) — the position and power level of the i-th beacon respectively. No two beacons will have the same position, so a_i ≠ a_j if i ≠ j.

Output

Print a single integer — the minimum number of beacons that could be destroyed if exactly one beacon is added.

Example

Input

4
1 9
3 1
6 1
7 4

Output

1

Input

7
1 1
2 1
3 1
4 1
5 1
6 1
7 1

Output

3

Note

For the first sample case, the minimum number of beacons destroyed is 1. One way to achieve this is to place a beacon at position 9 with power level 2.

For the second sample case, the minimum number of beacons destroyed is 3. One way to achieve this is to place a beacon at position 1337with power level 42.

其实就是要先sort一次，每次进行二分统计统计其level，但是要输出最小的，找到最大的减一下就行

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
int n,b[N];
pair<int,int>a[N];
int main()
{
    scanf("%d",&n);
    for(int i=0; i<n; i++)
        scanf("%d%d",&a[i].first,&a[i].second);
    sort(a,a+n);
    for(int i=0; i<n; i++)
    {
        int t=lower_bound(a,a+n,make_pair(a[i].first-a[i].second,0))-a;
        if(t) b[i]=b[t-1]+1;
        else b[i]=1;
    }
    printf("%d
",n-*max_element(b,b+n));
    return 0;
}

B - Bear and Prime Numbers

 CodeForces - 385C

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

You are given a sequence of integers x₁, x₂, ..., x_n of length n and mqueries, each of them is characterized by two integers l_i, r_i. Let's introduce f(p) to represent the number of such indexes k, that x_k is divisible by p. The answer to the query l_i, r_i is the sum: , where S(l_i, r_i) is a set of prime numbers from segment [l_i, r_i] (both borders are included in the segment).

Help the bear cope with the problem.

Input

The first line contains integer n (1 ≤ n ≤ 10⁶). The second line contains n integers x₁, x₂, ..., x_n (2 ≤ x_i ≤ 10⁷). The numbers are not necessarily distinct.

The third line contains integer m (1 ≤ m ≤ 50000). Each of the following m lines contains a pair of space-separated integers, l_i and r_i (2 ≤ l_i ≤ r_i ≤ 2·10⁹) — the numbers that characterize the current query.

Output

Print m integers — the answers to the queries on the order the queries appear in the input.

Example

Input

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.

1. 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.
2. 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.
3. The third query comes l = 4, r = 4. As this interval has no prime numbers, then the sum equals 0.

这个题目不难的，给你n个数，m次查询，每个操作给你一个区间[l,r],求a[i]能被[l,r]内素数整除的总数。

 这个很像上次csa的一道题

#include <bits/stdc++.h>
using namespace std;
const int N=1e7+5;
int c[N],a[N],f[N];
int main()
{
    int n,m;
    scanf("%d",&n);
    for(int i=0,x; i<n; i++)
        scanf("%d",&x),c[x]++;
    for(int i=2; i<N; i++)
    {
        a[i]=a[i-1];
        if(f[i]) continue;
        a[i]+=c[i];
        for(int j=i+i; j<N; j+=i)
            f[j]=1,a[i]+=c[j];
    }
    scanf("%d",&m);
    while(m--)
    {
        int l,r;
        scanf("%d%d",&l,&r);
        printf("%d
",a[min(r,N-1)]-a[min(l-1,N-1)]);
    }
    return 0;
}

用一下二分，也进行前缀和优化

#include <bits/stdc++.h>
using namespace std;
const int N=1e7+5;
int c[N],dp[N],a[N],f[N];
int main()
{
    int n,m,l=0,ma=0;
    scanf("%d",&n);
    for(int i=0,x; i<n; i++)
        scanf("%d",&x),a[x]++,ma=max(ma,x);
    for(int i=2; i<=ma; i++)
    {
        if(!f[i])
        {
            c[l++]=i;
            for(int j=i+i; j<=ma; j+=i)f[j]=1;
        }
    }
    for(int i=0; i<l;i++)
    {
        for(int j=c[i];j<=ma;j+=c[i])
            dp[i]+=a[j];
        if(i)dp[i]+=dp[i-1];
    }
    n=l;
    scanf("%d",&m);
    while(m--)
    {
        int l,r,x,y;
        scanf("%d%d",&l,&r);
        x=lower_bound(c,c+n,l)-c;
        if(r<=ma)y=upper_bound(c,c+n,r)-c;
        else y=n;
        printf("%d
",dp[y-1]-dp[x-1]);
    }
    return 0;
}