Codeforces 475D CGCDSSQ 求序列中连续数字的GCD=K的对数

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#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <iostream>
#include <cmath>
using namespace std;
typedef long long ll;
template <class T>
inline bool rd(T &ret) {
char c; int sgn;
if(c=getchar(),c==EOF) return 0;
while(c!='-'&&(c<'0'||c>'9')) c=getchar();
sgn=(c=='-')?-1:1;
ret=(c=='-')?0:(c-'0');
while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
ret*=sgn;
return 1;
}
template <class T>
inline void pt(T x) {
if(x>9) pt(x/10);
putchar(x%10+'0');
}
/////////////////////////
const int N = 100000 + 2;
struct Node {
int pos,l,r;
ll gval;
Node(int pos = 0,int l = 0,int r = 0,ll gval = 0):pos(pos),l(l),r(r),gval(gval){}
bool operator < (const Node & a) const {
if(gval != a.gval) return gval < a.gval;
if(pos != a.pos) return pos < a.pos;
return r < a.r;
}
}; int n, a[N], tot;
vector<Node> vt[N];
Node node[N * 50];
ll sum[N * 50];
void prepare() {
for(int i = 0;i <= n;++i) vt[i].clear();
vt[n].push_back(Node(n,n,n,a[n]));
Node ntmp;
int cnt, Size;
ll x;
for(int i = n - 1;i >= 1;--i) {
Size = vt[i + 1].size();
cnt = 1;
vt[i].push_back(Node(i,i,i,a[i]));
for(int j = 0;j < Size;++j) {
ntmp = vt[i+1][j];
x = __gcd((ll)a[i],ntmp.gval);
if(cnt && vt[i][cnt-1].gval == x)
vt[i][cnt - 1].r = ntmp.r;
else
vt[i].push_back(Node(i,ntmp.l,ntmp.r,x)),cnt++;
}
}
tot = 0;
for(int i = 1;i <= n;++i)
for(int j = 0;j < (int)vt[i].size();++j)
node[++tot] = Node(vt[i][j]);
sort(node + 1,node + tot + 1);
sum[0] = 0;
for(int i = 1;i <= tot;++i)
sum[i] = sum[i-1] + node[i].r - node[i].l + 1;
}
int hehe;
void work(int x) {
int L, R, l = 0, r = tot + 1, mid;
while (r - l > 1) {
mid = (l + r) >> 1;
if (node[mid].gval > x)
r = mid;
else
l = mid;
}
-- r;
if (r == 0 || node[r].gval != x) {
putchar('0');
putchar('\n');
return ;
}
R = r;
l = 0; r = tot + 1;
while (r - l > 1) {
mid = (l + r) >> 1;
if (node[mid].gval >= x)
r = mid;
else
l = mid;
}
L = r;
pt(sum[R] - sum[L - 1]);
putchar('\n');
}
int main() {
rd(n);
for (int i = 1; i <= n; ++i)
rd(a[i]);
prepare();
int Q, x;
rd(Q);
while (Q -- > 0) {
rd(x);
work(x);
}
return 0;
}
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