2015 Multi-University Training Contest 8 hdu 5381 The sum of gcd

The sum of gcd

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 641    Accepted Submission(s): 277

Problem Description

 You have an array A with the length of $n$

\[Let\quad f(l,r) = \sum_{i = l}^{r}\sum_{j = i}^{r}\quad gcd(a_{i},a_{i+1},\dots,a_{j})\]

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

First line has one integers $n$

Second line has $n$ integers $A_i$

Third line has one integers $Q$ the number of questions

Next there are $Q$ lines,each line has two integers $l,r$

$1 \leq T  \leq 3$

$1 \leq n,Q \leq 10^4$

$1 \leq a_i \leq 10^9$

$1\leq l,r \leq n$

 

Output

For each question,you need to print $f(l,r)$

 

Sample Input

2

5

1 2 3 4 5

3

1 3

2 3

1 4

4

4 2 6 9

3

1 3

2 4

2 3

 

Sample Output

9

6

16

18

23

10

 

Author

SXYZ

 

Source

2015 Multi-University Training Contest 8 1002

 

解题：线段树，学习了一位大神的写法，原来区间是可以这样子合并，本菜表示涨姿势了，again

 

先说说区间的合并吧，$gcd(gcd(1,2) gcd(3,4)) = gcd(1,4)$

 

我们先模拟下 四个数范围比较小的情况

 

1L:(11) 1R:(11)

2L:(22) 2R:(22)

3L:(33) 3R:(33)

4L:(44) 4R:(44)

 

1与2 合并 3与4合并

12L:(11,12) 12R:(22,12)

34L:(33,34) 34R:(44,34)

 

12 与 34 合并

1234L:(11,12,13,14)

1234R:(44,34,24,14)

非常巧妙。。。赞

每个R的最后一个都代表着整个区间

而且每个R的里面的区间最右边都是区间的最右边

所以左子树的右区间 与有子树的右区间是可以合并的

因为左子树的右区间都是以右子树的最左-1结尾的

 

同理 左区间合并。

 

至于数量的计算，上面说gcd的时候已经说了

 

另外，可能会有相同的gcd，所以用Ln Rn来记录相同gcd的出现次数

 

还有随着区间的增长，gcd不会变大，只会不变或者变得更小

 

 

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long LL;
 const int maxn = ;
 struct node {
     LL sum;
     int Lg[],Rg[],Ln[],Rn[],Lc,Rc;
 } tree[maxn<<];
 node pushup(const node &a,const node &b) {
     node ret;
     ret.sum = a.sum + b.sum;
     for(int i = ; i < a.Rc; ++i)
         for(int j = ; j < b.Lc; ++j)
             ret.sum += __gcd(a.Rg[i],b.Lg[j])*1LL*a.Rn[i]*b.Ln[j];
     for(int i = ; i < b.Rc; ++i) {
         ret.Rg[i] = b.Rg[i];
         ret.Rn[i] = b.Rn[i];
     }
     for(int i = ; i < a.Lc; ++i) {
         ret.Lg[i] = a.Lg[i];
         ret.Ln[i] = a.Ln[i];
     }
     int d = b.Rg[b.Rc-],p = b.Rc;
     for(int i = ; i < a.Rc; ++i) {
         int tmp = __gcd(a.Rg[i],d);
         if(tmp == ret.Rg[p-]) ret.Rn[p-] += a.Rn[i];
         else {
             ret.Rg[p] = tmp;
             ret.Rn[p++] = a.Rn[i];
         }
     }
     ret.Rc = p;
     d = a.Lg[a.Lc-],p = a.Lc;
     for(int i = ; i < b.Lc; ++i) {
         int tmp = __gcd(d,b.Lg[i]);
         if(tmp == ret.Lg[p-]) ret.Ln[p-] += b.Ln[i];
         else {
             ret.Ln[p] = b.Ln[i];
             ret.Lg[p++] = tmp;
         }
     }
     ret.Lc = p;
     return ret;
 }
 void build(int L,int R,int v) {
     if(L == R) {
         tree[v].Lc = tree[v].Rc = ;
         tree[v].Ln[] = tree[v].Rn[] = ;
         scanf("%d",&tree[v].Lg[]);
         tree[v].sum = tree[v].Rg[] = tree[v].Lg[];
         return;
     }
     int mid = (L + R)>>;
     build(L,mid,v<<);
     build(mid+,R,v<<|);
     tree[v] = pushup(tree[v<<],tree[v<<|]);
 }
 void query(int L,int R,int lt,int rt,int v) {
     if(lt <= L && rt >= R) {
         if(lt == L) tree[] = tree[v];
         else tree[] = pushup(tree[],tree[v]);
         return;
     }
     int mid = (L + R)>>;
     if(lt <= mid) query(L,mid,lt,rt,v<<);
     if(rt > mid) query(mid+,R,lt,rt,v<<|);
 }
 int main() {
     int kase,n,m,x,y;
     scanf("%d",&kase);
     while(kase--) {
         scanf("%d",&n);
         build(,n,);
         scanf("%d",&m);
         while(m--) {
             scanf("%d%d",&x,&y);
             query(,n,x,y,);
             printf("%I64d\n",tree[].sum);
         }
     }
     return ;
 }