hdu5726 rmq+二分

GCD

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 3301 Accepted Submission(s): 1189

Problem Description

Give you a sequence of N(N≤100,000) integers : a1,…,an(0

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;
ll a[maxn];
ll dp[maxn][30];
int Log[maxn];
int n,m;
ll gcd(ll x,ll y)
{
return x == 0?y:gcd(y%x,x);
}
void initRmq()
{
Log[0] = -1;
for(int i = 1; i <= n; i++)
{
Log[i] = (i&(i - 1)) == 0?Log[i - 1] + 1:Log[i - 1];
dp[i][0] = a[i];
}
for(int j = 1; j <= 20; 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 Rmp(int l,int r)
{
int x = Log[r - l + 1];
return gcd(dp[l][x],dp[r - (1<<x) + 1][x]);
}
ll bi_search(ll x,int i)
{
int l = i;
int r = n;
int result = l;
while(l <= r)
{
int mid = (l + r)>>1;
ll term = Rmp(i,mid);
if(term == x)
{
l = mid + 1;
result = mid;
}
else r = mid - 1;;
}
return result;
}
int main()
{
int T;
scanf("%d",&T);
int Case = 1;
while(T--)
{
map<ll,ll> num;
scanf("%d",&n);
for(int i = 1; i <= n; i++)
{
scanf("%I64d",&a[i]);
}
initRmq();
for(int i = 1; i <= n; i++)
{
ll x = a[i];
ll loc = i;
while(loc <= n)
{
ll loc1 = bi_search(x,loc);
//cout<<loc1<<endl;
num[x] += loc1 - loc + 1;
loc1++;
loc = loc1;
x = gcd(x,a[loc]);
}
}
int l,r;
scanf("%d",&m);
printf("Case #%d:\n",Case++);
for(int i = 1; i <= m; i++)
{
scanf("%d%d",&l,&r);
printf("%I64d %I64d\n",Rmp(l,r),num[Rmp(l,r)]);
}
}
return 0;
}