【题意】一个数列，对于每个询问f(l,r)求
【分析】把每个查询范围画出来就是一个倒三角，然后可以用莫队算法，关键就是单步转移；因为对于任意一个数已知gcd最多只有log(ai)种数字，所以用st预处理出每个点往右和往左求出的一段gcd值最多只有log(ai)种；这样就可以用log(ai)的复杂度转移，莫队复杂度就是q*sqrt(n)*log(ai)；而前面预处理需要二分来加速，这样配合st就是n*logn的(开始用线段树n*logn*logn预处理无线T，还是线段树常数太大)。
【AC CODE】390ms
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
#define rep(i,a,n) for(int i = a; i < n; i++)
#define repe(i,a,n) for(int i = a; i <= n; i++)
#define per(i,n,a) for(int i = n; i >= a; i--)
#define clc(a,b) memset(a,b,sizeof(a))
const int INF = 0x3f3f3f3f, MAXN = 10000+10, MAXIN = 10000,MAXOUT = 10000;
char buf[MAXIN], *ps = buf, *pe = buf+1;
inline void rnext(){
if(++ps == pe) pe = (ps = buf)+fread(buf,sizeof(char),sizeof(buf)/sizeof(char),stdin);
}
template <class T>
inline bool in_int(T &ans)
{
ans = 0;
T f = 1;
if(ps == pe) return false;
do{ rnext(); if('-' == *ps) f = -1;} while(!isdigit(*ps) && ps != pe);
if(ps == pe) return false;
do{ ans = (ans<<1)+(ans<<3)+*ps-48;rnext();}while(isdigit(*ps) && ps != pe);
ans *= f;
return true;
}
char bufout[MAXOUT], outtmp[50],*pout = bufout, *pend = bufout+MAXOUT;
inline void write(){ fwrite(bufout,sizeof(char),pout-bufout,stdout);pout = bufout;}/*必须在程序结束时加 write()*/
inline void out_char(char c){ *(pout++) = c;if(pout == pend) write();}
inline void out_str(char *s)
{
while(*s){ *(pout++) = *(s++); if(pout == pend) write(); }
}
template <class T>
inline void out_int(T x) {
if(!x){ out_char('0');return;}
if(x < 0) x = -x,out_char('-');
int len = 0;
while(x){ outtmp[len++] = x%10+48; x /= 10;}
outtmp[len] = 0;
for(int i = 0, j = len-1; i < j; i++,j--) swap(outtmp[i],outtmp[j]);
out_str(outtmp);
}
int a[MAXN],n;
int gcd(int a, int b)
{
while(b)
{
a %= b;
if(a < b) swap(a,b);
}
return a;
}
int dp[MAXN][20];
void st_init(int n)
{
repe(i,1,n) dp[i][0] = a[i-1];//下标从1开始的
for(int j = 1; (1<<j) <= n; j++)
{
for(int i = 1; i+(1<<j)-1 <= n; i++)//下标从1开始的
dp[i][j] = gcd(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
}
}
int st(int x, int y)
{
x++,y++;
int k = 0;
while((1<<(k+1)) <= y-x+1) k++;
return gcd(dp[x][k], dp[y-(1<<k)+1][k]);
}
struct NODE{
int x,y,id,block;
bool operator<(const NODE&t)const{
if(block != t.block) return block < t.block;
return y < t.y;
}
}in[MAXN];
struct P{
int len;
LL v[31],cnt[31];
}p[2][MAXN];
int find_r(int v, int l)
{
int x = l, y = n-1;
while(x <= y)
{
int m = (x+y)>>1;
if(st(l,m) >= v) x = m+1;
else y = m-1;
}
return x;
}
int find_l(int v, int r)
{
int x = 0,y = r;
while(x <= y)
{
int m = (x+y)>>1;
if(st(m,r) >= v) y = m-1;
else x = m+1;
}
return y;
}
LL ans[MAXN],sum;
void update(int tag, bool add, int st, int ed)
{
int cnt = ed-st+1,tol = 0;
P *now = &p[tag][st];
if(tag) now = &p[tag][ed];
while(cnt)
{
if(cnt >= now->cnt[tol])
{
if(add) sum += now->cnt[tol]*now->v[tol];
else sum -= now->cnt[tol]*now->v[tol];
cnt -= now->cnt[tol++];
}
else
{
if(add) sum += now->v[tol]*cnt;
else sum -= now->v[tol]*cnt;
cnt = 0;
}
}
}
int main()
{
#ifdef SHY
freopen("d:\\1.txt", "r", stdin);
freopen("d:\\out.txt", "w", stdout);
#endif
int t;
in_int(t);
while(t--)
{
in_int(n);
rep(i,0,n) in_int(a[i]);
st_init(n);
rep(i,0,n)
{
int pos = i,la = i,v = a[i];
p[0][i].len = 0;
while(1)
{
pos = find_r(v,i);
p[0][i].v[p[0][i].len] = v;
p[0][i].cnt[p[0][i].len++] = pos-la;
if(pos > n-1) break;
v = st(i,pos);
la = pos;
}
}
rep(i,0,n)
{
int pos = i,la = i,v = a[i];
p[1][i].len = 0;
while(1)
{
pos = find_l(v,i);
p[1][i].v[p[1][i].len] = v;
p[1][i].cnt[p[1][i].len++] = la-pos;
if(pos < 0) break;
v = st(pos,i);
la = pos;
}
}
int q, SIZE = sqrt(n+0.5);
in_int(q);
rep(i,0,q)
{
in_int(in[i].x), in_int(in[i].y);
in[i].x--;in[i].y--;
in[i].id = i;
in[i].block = in[i].x/SIZE;
}
sort(in,in+q);
sum = a[0];
int x = 0,y = 0;
rep(i,0,q)
{
while(x > in[i].x)
{
update(0,1,--x,y);
//sum += add[0][--x];
}
while(y < in[i].y)
{
update(1,1,x,++y);
//sum += add[1][++y];
}
while(x < in[i].x)
{
update(0,0,x++,y);
//sum -= add[0][x++];
}
while(y > in[i].y)
{
update(1,0,x,y--);
//sum -= add[1][y--];
}
ans[in[i].id] = sum;
}
rep(i,0,q) out_int(ans[i]),out_char('\n');
}
write();
return 0;
}