HDU 4777 Rabbit Kingdom （2013杭州赛区1008题，预处理，树状数组）

Rabbit Kingdom

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 40 Accepted Submission(s): 20


[align=left]Problem Description[/align]
　　Long long ago, there was an ancient rabbit kingdom in the forest. Every rabbit in this kingdom was not cute but totally pugnacious, so the kingdom was in chaos in season and out of season.
　　n rabbits were numbered form 1 to n. All rabbits' weight is an integer. For some unknown reason, two rabbits would fight each other if and only if their weight is NOT co-prime.
　　Now the king had arranged the n rabbits in a line ordered by their numbers. The king planned to send some rabbits into prison. He wanted to know that, if he sent all rabbits between the i-th one and the j-th one(including the i-th one and the j-th one) into prison, how many rabbits in the prison would not fight with others.
　　Please note that a rabbit would not fight with himself.

[align=left]Input[/align]
　　The input consists of several test cases.
　　The first line of each test case contains two integer n, m, indicating the number of rabbits and the queries.
　　The following line contains n integers, and the i-th integer Wi indicates the weight of the i-th rabbit.
　　Then m lines follow. Each line represents a query. It contains two integers L and R, meaning the king wanted to ask about the situation that if he sent all rabbits from the L-th one to the R-th one into prison.
　　(1 <= n, m, Wi <= 200000, 1 <= L <= R <= n)
　　The input ends with n = 0 and m = 0.

[align=left]Output[/align]
　　For every query, output one line indicating the answer.

[align=left]Sample Input[/align]

3 2
2 1 4
1 2
1 3
6 4
3 6 1 2 5 3
1 3
4 6
4 4
2 6
0 0

[align=left]Sample Output[/align]

2
1
1
3
1
2

Hint

　　In the second case, the answer of the 4-th query is 2, because only 1 and 5 is co-prime with other numbers in the interval [2,6] .

[align=left]Source[/align]
2013 Asia Hangzhou Regional Contest

关键是在预处理，每个数预处理出L,R区间，表示左右和这个数不互质的位置。

这个只要从左到右和从右到左扫描一遍，分解质因素，找下一个质因素的位置。

然后对于每个查询进行离线处理，按照右端点排序。

遇到i,在L处+1, 遇到R,在i处+1,在L处-1.

/* ***********************************************
Author        :kuangbin
Created Time  :2013-11-9 14:38:41
File Name     :E:\2013ACM\专题强化训练\区域赛\2013杭州\1008.cpp
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

const int MAXN = 200010;
int prime[MAXN+1];
void getPrime()
{
memset(prime,0,sizeof(prime));
for(int i = 2;i <= MAXN;i++)
{
if(!prime[i])prime[++prime[0]] = i;
for(int j = 1;j <= prime[0] && prime[j] <= MAXN/i;j++)
{
prime[prime[j]*i] = 1;
if(i % prime[j] == 0)break;
}
}
}
long long factor[100][2];
int fatCnt;
int getFactors(long long x)
{
fatCnt = 0;
long long tmp = x;
for(int i = 1;prime[i] <= tmp/prime[i];i++)
{
factor[fatCnt][1] = 0;
if(tmp % prime[i] == 0)
{
factor[fatCnt][0] = prime[i];
while(tmp % prime[i] == 0)
{
factor[fatCnt][1]++;
tmp /= prime[i];
}
fatCnt++;
}
}
if(tmp != 1)
{
factor[fatCnt][0] = tmp;
factor[fatCnt++][1] = 1;
}
return fatCnt;
}
int L[MAXN],R[MAXN];
int a[MAXN];
int b[MAXN];
int n,m;
int lowbit(int x)
{
return x & (-x);
}
int c[MAXN];
void add(int i,int val)
{
if(i == 0)return;
while(i <= n)
{
c[i] += val;
i += lowbit(i);
}
}
int sum(int i)
{
int s = 0;
while(i > 0)
{
s += c[i];
i -= lowbit(i);
}
return s;
}
vector<int>vec[MAXN];
struct Node
{
int l,r;
int index;
void input()
{
scanf("%d%d",&l,&r);
}
};
bool cmp(Node p1,Node p2)
{
return p1.r < p2.r;
}
Node node[MAXN];
int ans[MAXN];
int pp[MAXN][15];
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
getPrime();
while(scanf("%d%d",&n,&m) == 2)
{
if(n == 0 && m == 0)break;
for(int i = 1;i <= n;i++)
scanf("%d",&a[i]);
for(int i = 0;i < m;i++)
{
node[i].input();
node[i].index = i;
}
for(int i = 1;i < MAXN;i++)b[i] = n+1;
for(int i = n;i >= 1;i--)
{
getFactors(a[i]);
R[i] = n+1;
pp[i][0] = fatCnt;
for(int j = 0;j < fatCnt;j++)
{
R[i] = min(R[i],b[factor[j][0]]);
b[factor[j][0]] = i;
pp[i][j+1] = factor[j][0];
}
}
for(int i = 1;i < MAXN;i++)b[i] = 0;
for(int i = 1;i <= n;i++)
{
//getFactors(a[i]);
L[i] = 0;
fatCnt = pp[i][0];
for(int j = 0;j < fatCnt;j++)
{
factor[j][0] = pp[i][j+1];
L[i] = max(L[i],b[factor[j][0]]);
b[factor[j][0]] = i;
}
}
sort(node,node+m,cmp);
memset(c,0,sizeof(c));
for(int i = 1; i <= n+1;i++)
{
c[i] = 0;
vec[i].clear();
}
for(int i = 1;i <= n;i++)vec[R[i]].push_back(i);
int id = 1;
for(int i = 0;i < m;i++)
{
while(id <= n && id <= node[i].r)
{
add(L[id],1);
int sz = vec[id].size();
for(int j = 0;j < sz;j++)
{
int v = vec[id][j];
add(L[v],-1);
add(v,1);
}
id++;
}
ans[node[i].index] = sum(node[i].r) - sum(node[i].l-1);
ans[node[i].index] = node[i].r - node[i].l +1 - ans[node[i].index];
}
for(int i = 0;i < m;i++)printf("%d\n",ans[i]);

}
return 0;
}