HDU 5726 GCD

GCD

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 958    Accepted Submission(s): 288

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 forgcd(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

大致题意就是，给出一串数列，问你L到R之间的GCD，并要求出整个串中有几个子串的DCG和L到R的GCD相同。

用ST数组存放区间GCD值，用MAP存放整个序列中GCD为I的子串个数。

从L起始，向右延伸的子串，其GCD值都是单调非增的、所以可以枚举起点用二分来快速找到每个GCD对应的区间。

而数的范围是1e9，所以每个数的质因子个数最多有log(1e9)个

#include <iostream>#include <cstdio>#include <cstring>#include <map>#define LL long longusing namespace std;int st[100000+500][20];int num[100000+500];int n,m;map<int,LL>mp;int __gcd(int a,int b){    while(b)    {        int t=a%b;        a=b;        b=t;    }    return a;}void init(){    for(int i=1;i<=n;i++) st[i][0]=num[i];    for(int j=1;(1<<j)<=n;j++)    {        for(int i=1;i+(1<<j)-1<=n;i++)        {            st[i][j]=__gcd(st[i][j-1],st[i+(1<<(j-1))][j-1]);        }    }}int find_gcd(int l,int r){    int i=0;    while((1<<(i+1))<=r-l+1) i++;    return __gcd(st[l][i],st[r-(1<<i)+1][i]);}int main(){    int t,tt=1;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=1;i<=n;i++)        scanf("%d",&num[i]);        init();        scanf("%d",&m);        for(int i=1;i<=m;i++)        {            int l,r;            scanf("%d%d",&l,&r);            if(l>r) swap(l,r);            num[i]=find_gcd(l,r);            mp[num[i]]=0;        }        for(int i = 1;i<=n;i++)        {            int np = i;            while(np<=n)            {                int sum = find_gcd(i,np);                int l = np,r = n;                int cnt = 0;                while(l<=r)                {                    int mid = (l+r)/2;                    if(find_gcd(i,mid)==sum)                    {                        l = mid+1;                        cnt = mid;                    }                    if(find_gcd(i,mid)<sum)                       r = mid-1;                    else                       l = mid+1,cnt=mid;                }                mp[sum]+=cnt-np+1;                np = r+1;            }        }        printf("Case #%d:\n",tt++);        for(int i=1;i<=m;i++)        {            printf("%d %lld\n",num[i],mp[num[i]]);        }    }    return 0;}