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| #include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> PII;
typedef long long LL;
const int N = 100010, M = 18;
int n, m;
int a[N];
int f[N][M];
int Log2[N];
unordered_map<int, LL> ans;
int gcd(int a, int b) {
return b ? gcd(b, a % b) : a;
}
void ST_pre() {
Log2[2] = 1;
for (int i = 3; i < N; i++) Log2[i] = Log2[i >> 1] + 1;
for (int i = 1; i <= n; i++) f[i][0] = a[i];
for (int j = 1; (1 << j) < n; j++){
for (int i = 1; i + (1 << j) - 1 <= n; i++) {
f[i][j] = gcd(f[i][j - 1], f[i + (1 << j - 1)][j - 1]);
}
}
}
int query(int l, int r) {
int k = Log2[r - l + 1];
return gcd(f[l][k], f[r - (1 << k) + 1][k]);
}
// 在[st, ed]区间找到一个最大的k,gcd(left, k)=g
int find(int left, int st, int ed, int g) {
int l = st, r = ed;
while (l < r) {
int mid = l + r + 1 >> 1;
if (query(left, mid) >= g) l = mid;
else r = mid - 1;
}
return l;
}
void init() {
// 枚举区间左端点
for (int l = 1; l <= n; l++) {
int r = l;
while (r <= n) {
int g = query(l, r);
// 二分求以l为左端点,且gcd为g的最远右端点nextr∈[r,n]
int nextr = find(l, r, n, g);
// [r, nextr]这段的gcd为g
ans[g] += nextr - r + 1;
r = nextr + 1;
}
}
}
int main() {
cin >> n;
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
ST_pre();
init();
for (cin >> m; m--; ) {
int x; scanf("%d", &x);
cout << ans[x] << endl;
}
return 0;
}
|