區間修改單點查詢，又觀察到是一個k小，考慮主席樹上做差分
一開始樣例瘋狂掛，後來發現主席樹在一個歷史版本上只能修改一次，所以要開2*n個根結點，記錄一下每個時間對應的根結點編號
然後80分，考慮到當一個排名的結點有w個而查詢的k<w時會使答案變大，所以特判（但是一開始又喜聞樂見地把符號寫反了）~
一通亂改+瘋狂提交後水過

Code：

#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long ll;

const int
 
 N = 1e5 + 5;
const ll inf = (ll)1 << 60;

int n, qn, bel[N], maxn = 0;
ll val[N];
vector <int> ins[N], del[N];

struct Task {
    int st, ed;
    ll pri;
} a[N];

struct Innum {
    int id;
    ll val;
} in[N];

bool cmp(const Innum &x, const Innum &y) {
    if(x.val != y.val) return
 
 x.val < y.val;
    else return x.id < y.id;
}

template <typename T>
inline void read(T &X) {
    X = 0;
    char ch = 0;
    T op = 1;
    for(; ch > ‘9‘|| ch < ‘0‘; ch = getchar())
        if(ch == ‘-‘) op = -1;
    for(; ch >= ‘0‘ && ch <= ‘9‘; ch = getchar())
        X 
 
= (X << 3) + (X << 1) + ch - 48;
    X *= op;
}

inline int max(int x, int y) {
    return x > y ? x : y;
}

inline void discrete() {
    in[0].val = -inf;
    sort(in + 1, in + 1 + n, cmp);
    int cnt = 0;
    for(int i = 1; i <= n; i++) {
        if(in[i].val != in[i - 1].val) ++cnt;
        maxn = max(maxn, cnt);
        val[cnt] = in[i].val;
        a[in[i].id].pri = cnt;
    }
} 

struct Node {
    int lc, rc, cntSum;
    ll priSum;
};

namespace PSegT {
    int root[N << 1], nodeCnt;
    Node s[N * 40];
    
    #define mid (l + r) / 2
    
    inline void up(int p) {
        s[p].cntSum = s[s[p].lc].cntSum + s[s[p].rc].cntSum;
        s[p].priSum = s[s[p].lc].priSum + s[s[p].rc].priSum;
    }
        
    void insert(int &p, int l, int r, int x, int pre, int type) {
        p = ++nodeCnt;
        s[p] = s[pre];
        s[p].cntSum += type;
        s[p].priSum += (ll)type * val[x];

        if(l == r) return;
        
        if(x <= mid) insert(s[p].lc, l, mid, x, s[pre].lc, type);
        else insert(s[p].rc, mid + 1, r, x, s[pre].rc, type);
        
//        up(p);
    }
    
    ll query(int p, int l, int r, int k) {
//        if(l == r) return s[p].priSum;
        if(l == r) return k < s[p].cntSum ? (ll)k * val[l] : s[p].priSum;
        int now = s[s[p].lc].cntSum; ll res = 0;
        if(k <= now) res = query(s[p].lc, l, mid, k);
        else res = s[s[p].lc].priSum + query(s[p].rc, mid + 1, r, k - now);
        return res;
    }
    
    void print(int p, int l, int r) {
        printf("(%d %d %d %d %lld) ", p, s[p].lc, s[p].rc, s[p].cntSum, s[p].priSum);
        if(l == r) return;
        print(s[p].lc, l, mid);
        print(s[p].rc, mid + 1, r);
    }
    
    #undef mid
    
} using namespace PSegT;

int main() {
    read(n), read(qn);
    for(int i = 1; i <= n; i++) {
        read(a[i].st), read(a[i].ed), read(a[i].pri);
        in[i].val = a[i].pri, in[i].id = i;
        ins[a[i].st].push_back(i);
        del[a[i].ed + 1].push_back(i); 
    }
    discrete();
    
    nodeCnt = root[0] = 0;
    int rtCnt = 0;
    for(int i = 1; i <= qn; i++) {
        for(unsigned int j = 0; j < ins[i].size(); j++) 
            ++rtCnt, insert(root[rtCnt], 1, maxn, a[ins[i][j]].pri, root[rtCnt - 1], 1);
        for(unsigned int j = 0; j < del[i].size(); j++) 
            ++rtCnt, insert(root[rtCnt], 1, maxn, a[del[i][j]].pri, root[rtCnt - 1], -1); 
        bel[i] = rtCnt;
    }
    
/*    for(int i = 1; i <= rtCnt; i++, printf("\n"))
        print(root[i], 1, maxn);   */
    
    ll ans = 1;
    for(int x, k, _a, _b, _c; qn--; ) {
        read(x), read(_a), read(_b), read(_c);
        k = 1 + ((ll)_a * ans + _b) % _c;
        if(k >= s[root[bel[x]]].cntSum) printf("%lld\n", ans = s[root[bel[x]]].priSum);
        else printf("%lld\n", ans = query(root[bel[x]], 1, maxn, k));
    }   
    
    return 0;
}

感覺vector還挺快的……

Luogu 3168 [CQOI2015]任務查詢系統