2016多校联赛1D GCD（HDU5726）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5726

016多校联合训练的同学们~ 

GCD

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1151    Accepted Submission(s): 354

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

题目大意：n个数，编号1-n，有q个查询，输出查询区间的gcd值和等于该gcd值的总区间个数

思路：先预处理一下f数组

f[i][j]=以i为起点的2^j个数   f[1][0]=gcd(a[1])   f[2][1]=fcd(a[2],a[3])...

可以知道 f[i][j]=gcd(f[i][j-1],f[i+(1<<j-1)][j-1]);

那么我们就可以轻松O（1）找到任意区间的gcd

比如找区间L,R   则可以这样算     int k=log2(R-L+1);          return gcd(f[L][k],f[R-(1<<k)+1][k]);

用mp[value]记录区间gcd的值为value的个数。

再计算区间gcd值的个数的时候

我们先枚举起点，二分终点（因为一个区间的gcd值是单调不递增的嘛），然后不断更新mp[value]

具体过程还是看代码吧：

#include<map>#include<cmath>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;typedef long long LL;const int maxn=100005;map<int,LL>mp;LL gcd(LL x,LL y){return x%y==0?y:gcd(y,x%y);}LL n,a[maxn],f[111111][55];void presolve()//预处理f数组,f[i][j]表示gcd(a[i],a[i+1]...a[i+2^j-1])也就是a[i] 为起点2^j个数的gcd {  for(int i=n;i>=1;i--)  {    for(int j=0;i+(1<<j)-1<=n;j++)    {      if(j==0)f[i][j]=a[i];      else f[i][j]=gcd(f[i][j-1],f[i+(1<<j-1)][j-1]);    }  }}LL getgcd(int L,int R)//O(1)查询区间gcd; {  int k=log2(R-L+1);  return gcd(f[L][k],f[R-(1<<k)+1][k]);}int main(){  int T;  scanf("%d",&T);  for(int t=1;t<=T;t++)  {    scanf("%lld",&n);    memset(f,0,sizeof(f));    for(int i=1;i<=n;i++)scanf("%lld",&a[i]);;    int q,l,r;    presolve();    for(int i=1;i<=n;i++)//枚举左端点     {      int start=i;      while(start<=n)      {        int head=start;        int tail=n;        int value=getgcd(i,start);        int cnt=0;        while(head<=tail)        {          int mid=(head+tail)>>1;          int temp=getgcd(i,mid);          if(temp==value)          {            head = mid+1;            cnt = mid;          }          if(temp<value)                      tail = mid-1;          else              head = mid+1,cnt=mid;                    }        mp[value]+=cnt-start+1;//每次算完更新数量         start=tail+1;      }    }    printf("Case #%d:\n",t);    scanf("%d",&q);    while(q--)    {      scanf("%d%d",&l,&r);      int ans=getgcd(l,r);      cout<<ans<<" "<<mp[ans]<<endl;    }    mp.clear();  }}