2018年长沙理工大学第十三届程序设计竞赛 I 连续区间的最大公约数

连续区间的最大公约数

思路：参照BZOJ 4488: [Jsoi2015]最大公约数脑补出的一个\(map\)套\(vector\)的写法，写起来比线段树短，运行时间比线段树快。

代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define LL long long
#define pb emplace_back
#define pii pair<int, int>

const int N = 1e5 + 5;
int T, n, q, a[N], L[N], R[N], d[N], cnt;
LL ans[N];
vector<int> vc[N], p[N*20];
vector<LL> sum[N*20];
map<int, int> mp, _p, mmp;
int main() {
    scanf("%d", &T);
    for(int cs = 1; cs <= T; ++cs) {
        scanf("%d", &n);
        for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
        scanf("%d", &q);
        for (int i = 1; i <= q; ++i) {
            scanf("%d %d", &L[i], &R[i]);
            vc[R[i]].pb(i);
        }
        cnt = 0;
        for (int i = 1; i <= n; ++i) {
            for (auto it : mp) {
                int g = __gcd(it.fi, a[i]);
                if(_p.find(g) == _p.end()) _p[g] = it.se;
            }
            if(_p.find(a[i]) == _p.end()) _p[a[i]] = i;
            mp = _p;
            _p.clear();
            vector<pii> tmp;
            for (auto it : mp) tmp.pb(it.se, it.fi);
            for (int j = 0; j < tmp.size(); ++j) {
                if(j+1 == tmp.size()) tmp[j].fi = i;
                else tmp[j].fi = tmp[j+1].fi-1;
                int now;
                if(mmp.find(tmp[j].se) == mmp.end()) {
                    mmp[tmp[j].se] = ++cnt;
                    now = cnt;
                }
                else now = mmp[tmp[j].se];
                p[now].pb(tmp[j].fi);
                sum[now].pb(sum[now].size()==0?tmp[j].fi:sum[now].back()+tmp[j].fi);
            }
            for (int x : vc[i]) {
                int l = L[x];
                int t = lower_bound(tmp.begin(), tmp.end(), pii{l, 0})-tmp.begin();
                d[x] = tmp[t].se;
                int pp = mmp[d[x]];
                t = lower_bound(p[pp].begin(), p[pp].end(), l)-p[pp].begin();
                if(t-1>=0) ans[x] = sum[pp].back()-sum[pp][t-1]-(sum[pp].size()-t)*1LL*(l-1);
                else ans[x] = sum[pp].back()-sum[pp].size()*1LL*(l-1);
            }
        }
        printf("Case #%d:\n", cs);
        for (int i = 1; i <= q; ++i) printf("%d %lld\n", d[i], ans[i]);
        for (auto it : mmp) p[it.se].clear(), sum[it.se].clear();
        mmp.clear();
        mp.clear();
        for (int i = 1; i <= n; ++i) vc[i].clear();
    }
    return 0;
}