[HDOJ5726]GCD（RMQ，二分）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5726

题意：给定数列，求区间[L,R]的GCD的值，并求出有多少个子区间满足和[L,R]的GCD相等。

RMQ预处理所有区间的GCD，枚举所有区间的左边界i，起初固定右边界j，二分枚举右边界j的最大值，使得[i,j]区间内的GCD不变，更新对应GCD的值，大小为j-i。

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int maxn = 100100;
int n, q;
LL a[maxn];
LL dp[maxn][30];
map<LL, LL> cnt;

LL gcd(LL x, LL y) {
  return y == 0 ? x : gcd(y, x%y);
}

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

LL query(int l, int r) {
  int j = 0;
  while(1 << (j + 1) <= r - l + 1) j++;
  return gcd(dp[l][j], dp[r-(1<<j)+1][j]);
}

inline bool scan_d(LL &num) {
    char in;bool IsN=false;
    in=getchar();
    if(in==EOF) return false;
    while(in!='-'&&(in<'0'||in>'9')) in=getchar();
    if(in=='-'){ IsN=true;num=0;}
    else num=in-'0';
    while(in=getchar(),in>='0'&&in<='9'){
        num*=10,num+=in-'0';
    }
    if(IsN) num=-num;
    return true;
}

int main() {
  //freopen("in", "r", stdin);
  int T, _ = 1;
  scanf("%d", &T);
  int l, r;
  while(T--) {
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
      scan_d(a[i]);
    }
    st(); cnt.clear();
    for(int i = 1; i <= n; i++) {
      int j = i;
      while(j <= n) {
        LL cur = query(i, j);
        int lo = j, hi = n;
        while(lo <= hi) {
          int mid = (lo + hi) >> 1;
          if(query(i, mid) >= cur) lo = mid + 1;
          else hi = mid - 1;
        }
        cnt[cur] += (LL)(hi - j + 1);
        j = hi + 1;
      }
    }
    printf("Case #%d:
", _++);
    scanf("%d", &q);
    while(q--) {
      scanf("%d %d", &l, &r);
      LL ret = query(l, r);
      cout << ret << " " << cnt[ret] << endl;
    }
  }
  return 0;
}