• HDU


    HDU - 5381

    首先我们可以观察出从一个点开始的连续gcd最多只有log个, 因为gcd每次最少除2, 

    然后我们暴力莫队转移就好啦。

    #include<bits/stdc++.h>
    #define LL long long
    #define LD long double
    #define ull unsigned long long
    #define fi first
    #define se second
    #define mk make_pair
    #define PLL pair<LL, LL>
    #define PLI pair<LL, int>
    #define PII pair<int, int>
    #define SZ(x) ((int)x.size())
    #define ALL(x) (x).begin(), (x).end()
    #define fio ios::sync_with_stdio(false); cin.tie(0);
    
    using namespace std;
    
    const int N = 1e4 + 7;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const int mod = 1e9 + 7;
    const double eps = 1e-8;
    const double PI = acos(-1);
    
    template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;}
    template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;}
    template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;}
    template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;}
    
    mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
    
    const int B = 130;
    
    int Log[N];
    struct ST {
        int dp[N][15]; int ty;
        void build(int n, int b[]) {
            for(int i = -(Log[0] = -1); i < N; i++)
                Log[i] = Log[i - 1] + ((i & (i - 1)) == 0);
            for(int i = 1; i <= n; i++) dp[i][0] = b[i];
            for(int j = 1; j <= Log[n]; j++)
                for(int i = 1; i + (1 << j) - 1 <= n; i++)
                    dp[i][j] = __gcd(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
        }
        inline int query(int x, int y) {
            if(x > y) return 0;
            int k = Log[y - x + 1];
            return __gcd(dp[x][k], dp[y - (1 << k) + 1][k]);
        }
    } rmq;
    
    int n, q, a[N];
    LL ans[N];
    LL tmp[N];
    
    vector<int> LP[N], RP[N];
    vector<int> LV[N], RV[N];
    
    struct Qus {
        int L, R, id;
        bool operator < (const Qus &rhs) const {
            if(L / B == rhs.L / B) return R < rhs.R;
            return L < rhs.L;
        }
    } qus[N];
    
    void init() {
        for(int i = 1; i <= n; i++) {
            LP[i].clear();
            RP[i].clear();
            LV[i].clear();
            RV[i].clear();
        }
    }
    
    int main() {
        int T; scanf("%d", &T);
        while(T--) {
            scanf("%d", &n);
            init();
            for(int i = 1; i <= n; i++) {
                scanf("%d", &a[i]);
            }
    
            rmq.build(n, a);
    
            for(int i = 1; i <= n; i++) {
                int p = i, low, high, mid, gcd, ret;
                while(p <= n) {
                    ret = p;
                    gcd = rmq.query(i, p);
                    RP[i].push_back(p);
                    RV[i].push_back(gcd);
                    low = p, high = n;
                    while(low <= high) {
                        mid = low + high >> 1;
                        if(rmq.query(i, mid) == gcd) ret = mid, low = mid + 1;
                        else high = mid - 1;
                    }
                    p = ret + 1;
                }
                RP[i].push_back(n + 1);
                RV[i].push_back(0);
            }
            for(int i = 1; i <= n; i++) {
                int p = i, low, high, mid, gcd, ret;
                while(p >= 1) {
                    ret = p;
                    gcd = rmq.query(p, i);
                    LP[i].push_back(p);
                    LV[i].push_back(gcd);
                    low = 1, high = p;
                    while(low <= high) {
                        mid = low + high >> 1;
                        if(rmq.query(mid, i) == gcd) ret = mid, high = mid - 1;
                        else low = mid + 1;
                    }
                    p = ret - 1;
                }
                LP[i].push_back(0);
                LV[i].push_back(0);
            }
            scanf("%d", &q);
            for(int i = 1; i <= q; i++) {
                scanf("%d%d", &qus[i].L, &qus[i].R);
                qus[i].id = i;
            }
    
            sort(qus + 1, qus + 1 + q);
    
            int l = 1, r = 0;
            LL now = 0;
            for(int i = 1; i <= q; i++) {
                int L = qus[i].L, R = qus[i].R, id = qus[i].id;
                while(r < R) {
                    r++;
                    for(int i = 0; i < SZ(LP[r]) - 1; i++) {
                        if(LP[r][i + 1] >= l) {
                            now += 1LL * LV[r][i] * (LP[r][i] - LP[r][i + 1]);
                        }
                        else {
                            now += 1LL * LV[r][i] * (LP[r][i] - l + 1);
                            break;
                        }
                    }
                }
                while(l > L) {
                    l--;
                    for(int i = 0; i < SZ(RP[l]) - 1; i++) {
                        if(RP[l][i + 1] <= r) {
                            now += 1LL * RV[l][i] * (RP[l][i + 1] - RP[l][i]);
                        }
                        else {
                            now += 1LL * RV[l][i] * (r - RP[l][i] + 1);
                            break;
                        }
                    }
                }
                while(r > R) {
                    for(int i = 0; i < SZ(LP[r]) - 1; i++) {
                        if(LP[r][i + 1] >= l) {
                            now -= 1LL * LV[r][i] * (LP[r][i] - LP[r][i + 1]);
                        }
                        else {
                            now -= 1LL * LV[r][i] * (LP[r][i] - l + 1);
                            break;
                        }
                    }
                    r--;
                }
                while(l < L) {
                    for(int i = 0; i < SZ(RP[l]) - 1; i++) {
                        if(RP[l][i + 1] <= r) {
                            now -= 1LL * RV[l][i] * (RP[l][i + 1] - RP[l][i]);
                        }
                        else {
                            now -= 1LL * RV[l][i] * (r - RP[l][i] + 1);
                            break;
                        }
                    }
                    l++;
                }
                ans[id] = now;
            }
            for(int i = 1; i <= q; i++) {
                printf("%lld
    ", ans[i]);
            }
        }
        return 0;
    }
    
    /*
    */
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  • 原文地址:https://www.cnblogs.com/CJLHY/p/11181709.html
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