The sum of gcd

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 854 Accepted Submission(s): 363

Problem Description

You have an array

A,the length of

A is

n
Let

f(l,r)=∑ri=l∑rj=igcd(ai,ai+1....aj)

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

Ai
Third line has one integers

Q,the number of questions
Next there are Q lines,each line has two integers

1≤T≤3

1≤n,Q≤104

1≤ai≤109

1≤l<r≤n

Output

For each question,you need to print

f(l,r)

Sample Input

Sample Output

最難的不是莫隊。。而是。。處理轉移關係。。神題

rmq可以用來預處理gcd也是第一次知道。。

利用gcd特殊性質優化

#include <bits/stdc++.h> 
using namespace std;
#define fi first
#define se second
const int maxn = 10010;
typedef pair<int, int> pii;
typedef long long ll;
int gcd[maxn][20], mm[maxn];
int n,q,a[maxn], unit;
vector<pii> lef[maxn], rig[maxn];
ll ans[maxn];
struct node{
    int l, r, id;
    bool operator<(const node & b) const{
        if( l/unit == b.l/unit) return r < b.r;
        return l/unit < b.l/unit;
    }
}Query[maxn];
void init(){
    mm[0] = -1;
    for(int i=1; i<=n; i++){
        mm[i] = (i&(i-1)) == 0?mm[i-1]+1:mm[i-1];
        gcd[i][0] = a[i];
    }
    for(int j=1; j<=mm[n]; j++)
        for(int i=1; i+(1<<j)-1<=n; i++)
            gcd[i][j] = __gcd(gcd[i][j-1], gcd[i+(1<<(j-1))][j-1]);
}
int query(int l, int r){
    int k = mm[r-l+1];
    return __gcd(gcd[l][k], gcd[r-(1<<k)+1][k]);
}
void go(){
    for(int i=1; i<=n; i++)
        lef[i].clear(), rig[i].clear();
    int ans;
    for(int i=1; i<=n; i++){
        int L = 1, R = i, g = a[i], mid;
        while(true){
            while(L<=R){
                mid = (L+R)>>1;
                if(query(mid, i) == g) R = mid-1, ans = mid;
                else L = mid + 1;
            }
            lef[i].push_back(make_pair(g, ans));
            if(ans == 1) break;
            R = ans - 1; L = 1; g = query(R, i);
        }
    }
    for(int i=1; i<=n; i++){
        int L = i, R = n, g = a[i], mid;
        while(true){
            while(L <= R){
                mid = (L+R+1)>>1;
                if(query(i, mid) == g) L = mid+1, ans = mid;
                else R = mid-1;
            }
            rig[i].push_back(make_pair(g, ans));
            if(ans == n) break;
            L = ans + 1, R = n, g = query(i, L);
        }
    }
}
ll LEFT(int x, int L){
    int p = x;
    ll ans = 0;
    int size = lef[x].size();
    for(int i=0; i<size; i++){
        int g = lef[x][i].first, pos = max(L, lef[x][i].second);
        ans += (ll)g*(ll)(p - pos + 1);
        p = pos - 1;
        if( p < L ) break;
    }
    return ans;
}
ll RIGHT(int x, int R){
    int p = x;
    ll ans = 0;
    int size = rig[x].size();
    for(int i=0; i<size; i++){
        int g = rig[x][i].fi, pos = min(R, rig[x][i].se);
        ans += (ll)g*(ll)(pos - p + 1);
        p = pos + 1;
        if( p > R) break;
    }
    return ans;
}
void work(){
    int L = 1;
    int R = 0;
    sort(Query+1, Query+q+1);
    ll cnt = 0;
    for(int i=1; i<=q; i++){
        while(R<Query[i].r)
            cnt += LEFT(++R, L);
        while(R>Query[i].r)
            cnt -= LEFT(R--, L);
        while(L<Query[i].l)
            cnt -= RIGHT(L++, R);
        while(L>Query[i].l)
            cnt += RIGHT(--L, R);
        ans[Query[i].id] = cnt;
    }
}
int main(){
    int T;
    cin>>T;
    while(T--){
        cin>>n;
        unit = sqrt(n);
        for(int i=1; i<=n; i++)
            scanf("%d", a+i);
        init();
        go();
        scanf("%d", &q);
        for(int i=1; i<=q; i++){
            scanf("%d %d", &Query[i].l, &Query[i].r);
            Query[i].id = i;
        }
        work();
        for(int i=1; i<=q; i++)
            printf("%I64d\n", ans[i]);
    }
    return 0;
}