Luogu 3168 [CQOI2015]任务查询系统

区间修改单点查询，又观察到是一个k小，考虑主席树上做差分
一开始样例疯狂挂，后来发现主席树在一个历史版本上只能修改一次，所以要开2*n个根结点，记录一下每个时间对应的根结点编号
然后80分，考虑到当一个排名的结点有w个而查询的k<w时会使答案变大，所以特判（但是一开始又喜闻乐见地把符号写反了）~
一通乱改+疯狂提交后水过

Code：

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

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

int n, qn, bel[N], maxn = ;
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 = ;
    char ch = ;
    T op = ;
    for(; ch > ''|| ch < ''; ch = getchar())
        if(ch == '-') op = -;
    for(; ch >= '' && ch <= ''; ch = getchar())
        X = (X << ) + (X << ) + ch - ;
    X *= op;
}

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

inline void discrete() {
    in[].val = -inf;
    sort(in + , in +  + n, cmp);
    int cnt = ;
    for(int i = ; i <= n; i++) {
        if(in[i].val != in[i - ].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 << ], nodeCnt;
    Node s[N * ];

    #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 + , 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 = ;
        if(k <= now) res = query(s[p].lc, l, mid, k);
        else res = s[s[p].lc].priSum + query(s[p].rc, mid + , 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 + , r);
    }

    #undef mid

} using namespace PSegT;

int main() {
    read(n), read(qn);
    for(int i = ; 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 + ].push_back(i);
    }
    discrete();

    nodeCnt = root[] = ;
    int rtCnt = ;
    for(int i = ; i <= qn; i++) {
        for(unsigned int j = ; j < ins[i].size(); j++)
            ++rtCnt, insert(root[rtCnt], , maxn, a[ins[i][j]].pri, root[rtCnt - ], );
        for(unsigned int j = ; j < del[i].size(); j++)
            ++rtCnt, insert(root[rtCnt], , maxn, a[del[i][j]].pri, root[rtCnt - ], -);
        bel[i] = rtCnt;
    }

/*    for(int i = 1; i <= rtCnt; i++, printf("\n"))
        print(root[i], 1, maxn);   */

    ll ans = ;
    for(int x, k, _a, _b, _c; qn--; ) {
        read(x), read(_a), read(_b), read(_c);
        k =  + ((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]], , maxn, k));
    }   

    return ;
}

感觉vector还挺快的……