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49 lines
1.0 KiB
49 lines
1.0 KiB
#include <bits/stdc++.h>
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using namespace std;
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const int N = 200010, M = 18;
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/*
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输入样例：
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5 3
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4 12 3 6 7
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1 3
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2 3
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5 5
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答案：
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1
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3
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7
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*/
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int n, q;
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int w[N];
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int f[N][M];
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// 预处理
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void rmq() {
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for (int j = 0; j < M; j++) // 注意是j在外层,类似于区间DP,长度在外层
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for (int i = 1; i + (1 << j) - 1 <= n; i++) // i(行)在内层
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if (j == 0) // base case 边界值
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f[i][j] = w[i];
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else
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f[i][j] = __gcd(f[i][j - 1], f[i + (1 << j - 1)][j - 1]);
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}
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// 查询区间最大值
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int query(int l, int r) {
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int len = r - l + 1;
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int k = log2(len);
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return __gcd(f[l][k], f[r - (1 << k) + 1][k]);
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}
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int main() {
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// 加快读入
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ios::sync_with_stdio(false), cin.tie(0);
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cin >> n >> q;
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for (int i = 1; i <= n; i++) cin >> w[i];
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rmq();
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while (q--) {
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int l, r;
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cin >> l >> r;
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printf("%d\n", query(l, r));
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}
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return 0;
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} |