HDU_5729_rmq+二分

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

rmq修改成gcd的，关键是找个数，用二分来找，刚开始理解了好久，因为每个区间内gcd是递减的，所以可以优化暴力枚举。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
using namespace std;

int a[],n,dp[][];
map<int,long long> mp;
int gcd(int a,int b)
{
    return b?gcd(b,a%b):a;
}

void rmq_init(int len)
{
    for(int i = ;i <= len;i++)
    {
        dp[i][] = a[i];
    }

    for(int j = ;(<<j) <= len;j++)
    {
        for(int i = ;i+(<<j)- <= len;i++)
        {
            dp[i][j] = gcd(dp[i][j-],dp[i+(<<(j-))][j-]);
        }
    }
}

int rmq_query(int l,int r)
{
    int k = (int)(log((double)(r-l+))/log(2.0));
    return gcd(dp[l][k],dp[r-(<<k)+][k]);
}

int main()
{
    int T;
    scanf("%d",&T);
    for(int z = ;z <= T;z++)
    {
        mp.clear();
        printf("Case #%d:\n",z);
        scanf("%d",&n);
        for(int i = ;i <= n;i++)   scanf("%d",&a[i]);
        rmq_init(n);
        for(int i = ;i <= n;i++)
        {
            int l = i,r = i;
            while(r <= n)
            {
                int ll = r,rr = n;
                int v = rmq_query(l,r);
                while(ll <= rr)
                {
                    int mid = (ll+rr)/;
                    if(rmq_query(l,mid) >= v)  ll = mid+;
                    else    rr = mid-;
                }
                mp[v] += ll-r;
                r = ll;
            }
        }
        int t;
        scanf("%d",&t);
        while(t--)
        {
            int l,r;
            scanf("%d%d",&l,&r);
            int ans = rmq_query(l,r);
            printf("%d %lld\n",ans,mp[ans]);
        }
    }
    return ;
}