HDU 5726 (RMQ 二分)

题目链接：点击这里

题意：每次询问一个区间的gcd，和这个gcd相同的区间的个数。

区间的gcd个数最多不会超过 log(max) 个， 而且gcd序列又是递减的，所以直接暴力枚举左端点，对右端点二分求出每一个值的区间扔到map里面统计一下就好了。

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define maxn 100005
#define mod 1000000007

int dp[maxn][21];
int a[maxn];
int n, q;
map <int, long long> gg;

int scan () {
    char ch=' ';
    while(ch<'0'||ch>'9')ch=getchar();
    int x=0;
    while(ch<='9'&&ch>='0')x=x*10+ch-'0',ch=getchar();
    return x;
}

void rmq_init () {
    for (int i = 0; i < n; i++) dp[i][0] = a[i];
    for (int j = 1; (1<for (int i = 0; i+(1<1 < n; i++) {
            dp[i][j] = __gcd (dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
        }
    }
}

int rmq (int l, int r) {
    int k = 0;
    while ((1<<(k+1)) <= r-l+1) k++;
    return __gcd (dp[l][k], dp[r-(1<1][k]);
}

void solve () {
    for (int i = 0; i < n; i++) {
        int pos = n-1;
        while (pos >= i) { 
            int tmp = rmq (i, pos);
            int l = i, r = pos;
            while (r-l > 1) {
                int mid = (l+r)>>1;
                if (rmq (i, mid) == tmp) r = mid;
                else l = mid;
            }
            int cur;
            if (rmq (i, l) == tmp) cur = l;
            else cur = r;
            gg[tmp] += (pos-cur+1);
            pos = cur-1;
        }
    }
}

int main () {
    int t, kase = 0;
    scanf ("%d", &t);
    while (t--) {
        printf ("Case #%d:\n", ++kase);
        gg.clear ();
        n = scan ();
        for (int i = 0; i < n; i++) a[i] = scan (); 
        rmq_init ();
        solve (); 
        q = scan ();
        while (q--) {
            int l = scan (), r = scan ();
            int ans = rmq (l-1, r-1);
            printf ("%d %lld\n", ans, gg[ans]);
        }
    }
    return 0;
}