HDU 5726 GCD (ST 表 +查询)

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=5726

GCD

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4341    Accepted Submission(s): 1550

Problem Description

Give you a sequence of N(N≤100,000) integers : a1,...,an(0<ai≤1000,000,000). There are Q(Q≤100,000) queries. For each query l,r you have to calculate gcd(al,,al+1,...,ar) and count the number of pairs(l′,r′)(1≤l<r≤N)such that gcd(al′,al′+1,...,ar′) equal gcd(al,al+1,...,ar).

 

Input

The first line of input contains a number T, which stands for the number of test cases you need to solve.

The first line of each case contains a number N, denoting the number of integers.

The second line contains N integers, a1,...,an(0<ai≤1000,000,000).

The third line contains a number Q, denoting the number of queries.

For the next Q lines, i-th line contains two number , stand for the li,ri, stand for the i-th queries.

 

Output

For each case, you need to output “Case #:t” at the beginning.(with quotes, t means the number of the test case, begin from 1).

For each query, you need to output the two numbers in a line. The first number stands for gcd(al,al+1,...,ar) and the second number stands for the number of pairs(l′,r′) such that gcd(al′,al′+1,...,ar′) equal gcd(al,al+1,...,ar).

 

Sample Input

151 2 4 6 741 52 43 44 4

 

Sample Output

Case #1:1 82 42 46 1

 

Author

HIT

 

Source

2016 Multi-University Training Contest 1

123

题意 :  给 n 个数  m 次查询 问 从 a-b  GCD是多少  并且问 你 在 整个n区间内 GCD = GCD(a,b)的 有多少个;

10w 的数据    QMR查询  利用ST  动态 打表, 求GCD

二分从 1-n  存取 GCD的相等的数目:

#include <iostream>#include <stdio.h>#include <algorithm>#include <cmath>#include <math.h>#include <cstring>#include <string>#include <queue>#include <stack>#include <stdlib.h>#include <list>#include <map>#include <set>#include <bitset>#include <vector>#define mem(a,b) memset(a,b,sizeof(a))#define findx(x) lower_bound(b+1,b+1+bn,x)-b#define FIN      freopen("input.txt","r",stdin)#define FOUT     freopen("output.txt","w",stdout)#define S1(n)    scanf("%d",&n)#define SL1(n)   scanf("%I64d",&n)#define S2(n,m)  scanf("%d%d",&n,&m)#define SL2(n,m)  scanf("%I64d%I64d",&n,&m)#define Pr(n)     printf("%d\n",n)using namespace std;typedef long long ll;const double PI=acos(-1);const int INF=0x3f3f3f3f;const double esp=1e-6;const int maxn=1e5+5;const int MOD=1e9+7;const int mod=1e9+7;int dir[5][2]={0,1,0,-1,1,0,-1,0};ll gcd(ll a,ll b){ return b?gcd(b,a%b):a;}ll exgcd(ll a,ll b,ll &x,ll &y){if(!b){x=1;y=0;return a;}ll ans=exgcd(b,a%b,x,y);ll temp=x;x=y;y=temp-a/b*y;return ans;}ll lcm(ll a,ll b){ return b/gcd(a,b)*a;}ll qpow(ll x,ll n){ll res=1;for(;n;n>>=1){if(n&1)res=(res*x)%MOD;x=(x*x)%MOD;}return res;}int n;ll a[maxn];ll maps[maxn][25];void ST(){    for(int i=1;i<=n;i++)        maps[i][0]=a[i];    for(int j=1;(1<<j)<=n;j++)        for(int i=1;i+(1<<j)-1<=n;i++)            maps[i][j]=gcd(maps[i][j-1],maps[i+(1<<j-1)][j-1]);}ll get_gcd(int l,int r){    if(l>r)swap(l,r);    int x= int (log(r-l+1)/log(2));    return gcd(maps[l][x],maps[r-(1<<x)+1][x]);}int main(){    int T;    S1(T);    int cont=0;    while(T--)    {        S1(n);        mem(a,0);        mem(maps,0);        for(int i=1;i<=n;i++)            S1(a[i]);        map<ll,ll>mp;        ST();        for(int i=1;i<=n;i++)        {             ll val= a[i];             int pos =i;             while(pos<=n)             {                 //cout<<"pos "<<pos<<endl;                int l=pos;                int r=n;                val=get_gcd(i,pos);                while(r>l)                {                    int mid=(l+r+1)>>1;                    if(get_gcd(i,mid)==val)                            l=mid;                     else                        r=mid-1;                 }                mp[val]+=(l-pos)+1;                pos=l+1;             }        }        int m;        printf("Case #%d:\n",++cont);        cin>>m;        int x,y;        while(m--)        {            scanf("%d %d",&x,&y);            ll ans=get_gcd(x,y);            printf("%lld %lld\n",ans,mp[ans]);        }    }    return 0;}