GCD (RMQ + 二分)

RMQ存的是区间GCD，然后遍历 i: 1->n, 然后不断地对[i, R]区间进行二分求以i为起点的相同gcd的区间范围，慢慢缩减区间。

#include<bits/stdc++.h>
#define LL long long
using namespace std;

const int maxn = 1e5 + ;
int in[maxn], dp[maxn][], mm[maxn];
map<int, LL>cnt;

void init(int n){
    mm[] = -;
    for(int i = ; i <= n; i ++){
        mm[i] = (i&(i - ))?mm[i - ]:mm[i - ] + ;
        dp[i][] = in[i];
    }
    for(int j = ; j <= mm[n]; j ++)
        for(int i = ; i + (<<j) -  <= n; i ++)
            dp[i][j] = __gcd(dp[i][j - ], dp[i + (<<(j-))][j - ]);
}

int RMQ(int l, int r){
    int k = mm[r - l + ];
    return __gcd(dp[l][k], dp[r - (<<k) + ][k]);
}

int main(){
    int T,n,m;scanf("%d",&T);
    for(int ncase = ; ncase <= T; ncase ++){
        printf("Case #%d:\n",ncase);
        scanf("%d",&n);
        for(int i = ; i <= n; i ++)scanf("%d",&in[i]);
        init(n);cnt.clear();
        int L, R, l, r, val, m;
        for(int i = ; i <= n; i ++){
            cnt[in[i]] ++;
            L = i;R = n;
            while(L < R){
                val = RMQ(L,R) + ;
                l = L;r = R;
                while(l < r){
                    m = (l + r)/;
                    if(RMQ(L, m) >= val)
                        l = m + ;
                    else
                        r = m;
                }
                cnt[val - ] += R - m;
                R = m;
            }
        }
        scanf("%d",&m);
        for(int i = ; i < m; i ++){
            scanf("%d%d",&l,&r);
            int val = RMQ(l,r);
            printf("%d %lld\n",val, cnt[val]);
        }
    }
    return ;
}