Hdoj 5381 The sum of GCD-mo team algorithm

The Great God:

http://blog.csdn.net/u014800748/article/details/47680899

The sum of GCDTime limit:2000/1000 MS (java/others) Memory limit:65536/65536 K (java/others)
Total submission (s): 526 Accepted Submission (s): 226


Problem Descriptionyou has an arrayThe length ofIs
Let

Inputthere is multiple test cases. The first line of input contains an integer T, indicating the number of the test cases. For each test case:
First line have one integers
Second Line hasIntegers
Third line have one integersThe number of questions
Next there is Q Lines,each line has both integers,




Outputfor each question,you need to print
Sample Input

251 2 3 4 531 32 31 444 2 6 931 32 42 3

Sample Output

9616182310

Authorsxyz
Source2015 multi-university Training Contest 8

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include < vector>using namespace Std;const int maxn=10100;typedef long Long int ll;struct g{g () {} g (int _id,ll _g): ID (_id) , g (_g) {} int id; LL g;void toString () {printf ("ID:%d g:%lld\n", id,g);};    int n,a[maxn],q;vector<g> vl[maxn],vr[maxn];struct que{int l,r,id; BOOL operator< (const que& Que) Const {if (l!=que. L) return L<que.        L Return R<que.    R }}que[maxn];void PreInit () {///Get left point////with I as the right endpoint, preprocess the segment for (int i=1;i<=n;i++) {vl[i].cle        AR ();        if (i==1) {vl[i].push_back (G (I,a[i])); } else {LL curg=a[i];int l=i;for (auto &it:vl[i-1]) {int G=__GCD (It.g,curg), if (G!=curg) Vl[i].push_back ( G (L,curg)); curg=g; L=it.id;}        Vl[i].push_back (G (L,curg)); }}///Get right point///with I as the left end, preprocess the segment for (int i=n;i>=1;i--) {vr[i].clear (), if (i==n) {Vr[i].push_Back (G (i,a[i));} Else{ll curg=a[i];int r=i;for (auto &it:vr[i+1]) {int G=__GCD (CURG,IT.G), if (G!=curg) Vr[i].push_back (g (R,curg)); curg=g; R=it.id;} Vr[i].push_back (G (R,curg));}}} Calculate the value between L,r ll calu (int type,int l,int R) {LL ret=0;if (type==0) {int tr=r;for (auto &it:vl[r]) {if (it.id>=l) {ret+= (tr-it.id+1) *it.g;tr=it.id-1;} else{ret+= (tr-l+1) *it.g;break;}} else if (type==1) {int tr=l;for (auto &it:vr[l]) {if (it.id<=r) {ret+= (it.id-tr+1) *it.g;tr=it.id+1;} else{ret+= (r-tr+1) *it.g;break;}} return ret;}    LL Ans[maxn];int Main () {int t_t;scanf ("%d", &t_t);        while (t_t--) {scanf ("%d", &n);        for (int i=1;i<=n;i++) scanf ("%d", a+i);        PreInit (); scanf ("%d", &q);            for (int i=0,l,r;i<q;i++) {scanf ("%d%d", &l,&r); Que[i]. L=l; Que[i]. R=r;        Que[i].id=i; } sort (que,que+q); int l=1,r=0; LL ret=0;for (int i=0;i<q;i++) {while (R<que[i]. R) {R++;ret+=calu (0,l,r);} while (R>que[i]. R) {Ret-=calu (0,l,r); r--;} while (L<que[i]. L) {Ret-=calu (1,l,r); l++;} while (L>que[i]. L) {L--;ret+=calu (1,l,r);} Ans[que[i].id]=ret;}    for (int i=0;i<q;i++) cout<<ans[i]<<endl; } return 0;}

Hdoj 5381 The sum of GCD-mo team algorithm