CGCDSSQ Codeforces 475D

思路：将数据存入gcd二维数组的第0列，第j列的gcd值是第j-1列的i行和i+2的j次方的最大公约数，

gcd[i][j] = Gcd(gcd[i][j - 1], gcd[i + bin[j - 1]][j - 1]);

将gcd数组计算好后，进行询问query，由于系统的Log2太慢，所以编写了Log数组直接使用。

int k = Log[r - l + 1];

return Gcd(gcd[l][k], gcd[r - bin[k] + 1][k]);
二分查找bi函数:将左端固定值为x，寻找值仍为x的右端，并返回；遍历完成即统计完成。

#include<iostream>
#include <cstdio>
#include <cstring>
#include<algorithm>
#include<map>
#define maxn 100010
using namespace std;
typedef long long LL;
int gcd[maxn][32], n, Log[maxn], bin[32];
map<int, LL> mp;
int Gcd(int a, int b) {
    return b ? Gcd(b, a % b) : a;
}
void rmqInit() {
    Log[0] = -1; bin[0] = 1;
    for (int i = 1; i < 20; ++i) {
        bin[i] = bin[i - 1] << 1;//1 2 4 8 16 32
    }
    for (int i = 1; i <= n; ++i) {
        Log[i] = Log[i >> 1] + 1;//-1 0 1 1 2 2 2 2 3 3 3 3 3 3 3 3 
    }
    for (int j = 1; bin[j] <= n; ++j) {
        for (int i = 1; i + bin[j - 1] - 1 <= n; ++i) {
            gcd[i][j] = Gcd(gcd[i][j - 1], gcd[i + bin[j - 1]][j - 1]);
        }
    }
}
int query(int l, int r) {//询问区间为l，r的gcd值
    int k = Log[r - l + 1];//
    return Gcd(gcd[l][k], gcd[r - bin[k] + 1][k]);//bin[k]就是2的k次方
}
int bi(int i, int l, int r, int x) {  //x是i到l的最大公约数
    while (r - l > 1) {  //右区间-左区间等于1时停止
        int mid = l + r >> 1;  
        int val = query(i, mid);
        if (val >= x) l = mid;   
        else r = mid - 1; 
    }
    //找出最大公约数是x的右区间
    return query(i, r) == x ? r : l;
}
int main() {
    scanf_s("%d", &n);
    for (int i = 1; i <= n; ++i) {
        cin >> gcd[i][0];
    }
    rmqInit();
    for (int i = 1; i <= n; ++i) {
        int l = i;    //左区间从1到n，一行一行搜索
        while (true) {
            if (l == n + 1) break;  //左区间超过n则跳出
            int val = query(i, l);   //询问区间为i，l的gcd值
            int r = bi(i, l, n, val); 
            mp[val] += r - l + 1;
            l = r + 1;
        }
    }
    int q, x;
    scanf_s("%d", &q);
    while (q--) {
        scanf_s("%d", &x);
        printf("%I64d\n", mp[x]);
    }
    return 0; 
}