Codeforces 837G Functions On The Segments 主席树

Functions On The Segments

考虑处理出所有x <= 2e5的答案。

用主席树去维护， 每个 x 的值， 因为是单点询问所以可以差分， 不用打标记。

一个PLL 打成了 PII， 找了半天bug。

#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 = 2e5 + 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;}

const int UP = 200000;

int Rt[N];
int treecnt;

struct info {
    LL a, b;
    int ls, rs;
    // a -> add val
    // b -> cnt of add i
} Tree[N * 60];

void update(int p, int va, int vb, int l, int r, int &x, int y) {
    x = ++treecnt; Tree[x] = Tree[y];
    Tree[x].a += va; Tree[x].b += vb;
    if(l == r) return;
    int mid = l + r >> 1;
    if(p <= mid) update(p, va, vb, l, mid, Tree[x].ls, Tree[y].ls);
    else update(p, va, vb, mid + 1, r, Tree[x].rs, Tree[y].rs);
}

PLL query(int L, int R, int l, int r, int x) {
    if(R < l || r < L || R < L) return mk(0, 0);
    if(L <= l && r <= R) return mk(Tree[x].a, Tree[x].b);
    int mid = l + r >> 1;
    PLL ans = query(L, R, l, mid, Tree[x].ls);
    PLL tmp = query(L, R, mid + 1, r, Tree[x].rs);
    ans.fi += tmp.fi;
    ans.se += tmp.se;
    return ans;
}

int n, m;
LL prefix[N];

int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
        int X1, X2, Y1, A, B, Y2, low, high;
        scanf("%d%d%d%d%d%d", &X1, &X2, &Y1, &A, &B, &Y2);
        prefix[i] = prefix[i - 1] + Y2;

        low = 0; high = min(UP, X1);
        update(low, Y1, 0, 0, UP, Rt[i], Rt[i - 1]);
        if(high + 1 <= UP) update(high + 1, -Y1, 0, 0, UP, Rt[i], Rt[i]);

        if(X1 + 1 <= UP) {
            low = X1 + 1; high = min(UP, X2);
            update(low, B, A, 0, UP, Rt[i], Rt[i]);
            if(high + 1 <= UP) update(high + 1, -B, -A, 0, UP, Rt[i], Rt[i]);
        }

        if(X2 + 1 <= UP) {
            low = X2 + 1; high = UP;
            update(low, Y2, 0, 0, UP, Rt[i], Rt[i]);
        }
    }

    LL last = 0;
    scanf("%d", &m);
    for(int cas = 1; cas <= m; cas++) {
        int L, R, x;
        scanf("%d%d%d", &L, &R, &x);
        x = (x + last % 1000000000) % 1000000000;
        if(x > UP) last = prefix[R] - prefix[L - 1];
        else {
            PLL tmpR = query(0, x, 0, UP, Rt[R]);
            PLL tmpL = query(0, x, 0, UP, Rt[L - 1]);
            last = (tmpR.fi - tmpL.fi) + (tmpR.se - tmpL.se) * x;
        }
        printf("%lld
", last);
    }
    return 0;
}

/*
*/