[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 = ;
 int n, q;
 LL a[maxn];
 LL dp[maxn][];
 map<LL, LL> cnt;

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

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

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

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

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