题目链接：http://codeforces.com/contest/837/problem/G

题意：给了n个分段函数，然后给m个查询，每次查询查[l,r]的f(x)的和，强制在线。

解法：关键是找到一个可以求前缀信息的数据结构，主席树显然满足，这题就是主席树的应用。


#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int mod = 1000000000;
const int N = 2e5+5;
struct node{
    int l,r;
    LL A,B,Y1,Y2;
}T[40*N];
int root[75010], cnt;
int n,m,x,l,r,AA,BB,YY1,YY2;
void pushup(int x){
    int L = T[x].l, R = T[x].r;
    T[x].A = T[L].A + T[R].A;
    T[x].B = T[L].B + T[R].B;
    T[x].Y1 = T[L].Y1 + T[R].Y1;
    T[x].Y2 = T[L].Y2 + T[R].Y2;
}
void update(int &x, int y, int l, int r, int pos)
{
    x = ++cnt, T[x] = T[y];
    if(l == r){
        T[x].A += AA, T[x].B += BB, T[x].Y1 += YY1, T[x].Y2 += YY2;
        return;
    }
    int mid = (l+r)/2;
    if(pos<=mid) update(T[x].l, T[y].l, l, mid, pos);
    else update(T[x].r, T[y].r, mid+1, r, pos);
    pushup(x);
}
LL query(int rt, int l, int r, int L, int R, int x){
    if(L <= l && r <= R){
        return T[rt].A*x + T[rt].B + T[rt].Y1 + T[rt].Y2;
    }
    int mid = (l+r)/2;
    if(R<=mid) return query(T[rt].l, l, mid, L, R, x);
    else if(L>mid) return query(T[rt].r, mid+1, r, L, R, x);
    else{
        return query(T[rt].l, l, mid, L, mid, x) + query(T[rt].r, mid+1, r, mid+1, R, x);
    }
}
int main()
{
    scanf("%d", &n);
    int x1, x2, y1, a, b, y2;
    for(int i=1; i<=n; i++){
        scanf("%d %d %d %d %d %d", &x1,&x2,&y1,&a,&b,&y2);
        AA = 0, BB = 0, YY1 = y1, YY2 = 0;
        update(root[i], root[i-1], 1, N, 1);
        AA = a, BB = b, YY1 = -y1, YY2 = 0;
        update(root[i], root[i], 1, N, x1+1);
        AA = -a, BB = -b, YY1 = 0, YY2 = y2;
        update(root[i], root[i], 1, N, x2+1);
    }
    scanf("%d", &m);
    LL ans = 0;
    while(m--)
    {
        scanf("%d %d %d", &l,&r,&x);
        x = (x + ans)%mod;
        LL t1 = query(root[r], 1, N, 1, x, x), t2 = query(root[l-1], 1, N, 1, x, x);
        ans = t1 - t2;
        printf("%lld\n", ans);
    }
    return 0;
}