【HDU5726 2016 Multi-University Training Contest 1D】【gcd的下降性质 STL-map】GCD 多少段区间gcd等于给定区间gcd

GCD

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

Total Submission(s): 2189    Accepted Submission(s): 731


[align=left]Problem Description[/align]
Give you a sequence of N(N≤100,000) integers
: a1,...,an(0<ai≤1000,000,000).
There are Q(Q≤100,000) queries.
For each query l,r you
have to calculate gcd(al,,al+1,...,ar) and
count the number of pairs(l′,r′)(1≤l<r≤N)such
that gcd(al′,al′+1,...,ar′) equal gcd(al,al+1,...,ar).
 

[align=left]Input[/align]
The first line of input contains a number T,
which stands for the number of test cases you need to solve.

The first line of each case contains a number N,
denoting the number of integers.

The second line contains N integers, a1,...,an(0<ai≤1000,000,000).

The third line contains a number Q,
denoting the number of queries.

For the next Q lines,
i-th line contains two number , stand for the li,ri,
stand for the i-th queries.

 

[align=left]Output[/align]
For each case, you need to output “Case #:t” at the beginning.(with quotes, t means
the number of the test case, begin from 1).

For each query, you need to output the two numbers in a line. The first number stands for gcd(al,al+1,...,ar) and
the second number stands for the number of pairs(l′,r′) such
that gcd(al′,al′+1,...,ar′) equal gcd(al,al+1,...,ar).

 

[align=left]Sample Input[/align]

1
5
1 2 4 6 7
4
1 5
2 4
3 4
4 4

 

[align=left]Sample Output[/align]

Case #1:
1 8
2 4
2 4
6 1

 

[align=left]Author[/align]
HIT
 

[align=left]Source[/align]
2016 Multi-University Training
Contest 1

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 1e5 + 10, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;
int casenum, casei;
int n, v
;
vector< pair<int,int> >lft
;
int gcd(int x, int y)
{
return y == 0 ? x : gcd(y, x%y);
}
map<int, LL>mop;
void STinit()
{
lft[n + 1].clear();
for (int i = n; i >= 1; i--)
{
lft[i].clear();
int p = i; int g = v[i];
for (auto& it : lft[i + 1])
{
int newg = gcd(g, it.second);
if (newg != g)lft[i].push_back(MP(p, g));
p = it.first;
g = newg;
}
lft[i].push_back(MP(p, g));
p = i - 1;
for (auto& it : lft[i])
{
mop[it.second] += it.first - p;
p = it.first;
}
}
}
int getgcd(int l, int r)
{
for (auto& it : lft[l])
{
if (r <= it.first)return it.second;
}
}
int main()
{
scanf("%d", &casenum);
for (casei = 1; casei <= casenum; ++casei)
{
mop.clear();
scanf("%d", &n); for (int i = 1; i <= n; ++i)scanf("%d", &v[i]);
STinit();
int q; scanf("%d", &q);
printf("Case #%d:\n", casei);
while (q--)
{
int l, r; scanf("%d%d", &l, &r);
int g = getgcd(l, r);
printf("%d %lld\n", g, mop[g]);
}
}
return 0;
}
/*
【trick&&吐槽】
比赛的时候把最后一个区间段忘记加进去了，导致吃了一发WA，悲剧！

【题意】
有n（1e5）个数，
每次给你一个区间段，
问你有多少个区间段内的gcd与该区间段的gcd是相同的，输出这样的区间个数。

【类型】
gcd的下降性质

【分析】
我们以一个点为区间的左界，然后逐渐扩展右界，gcd一旦变化了，至少也会变为原来的一半。
所以gcd的变化区间段数最多为logn段。
于是我们可以以区间的右侧开始，向右方向扩展gcd，然后存到map中。
这个预处理的复杂度是O(nlogn)
接下来对于单组数据的查询和询问，复杂度也是O(nlogn)，这道题就做完了。

【时间复杂度&&优化】
O(nlogn)

*/