Codeforces 311D Interval Cubing 数学 + 线段树 (看题解)

Interval Cubing

这种数学题谁顶得住啊。

因为 (3 ^ 48) % (mod - 1)为 1 , 所以48个一个循环节, 用线段树直接维护。

#include<bits/stdc++.h>
#define LL 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 ull unsigned long long

using namespace std;

const int N = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 95542721;
const double eps = 1e-6;
const double PI = acos(-1);

int n, q;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
int a[N << 2][48], lazy[N << 2];

inline void pull(int rt) {
    for(int i = 0; i < 48; i++) {
        a[rt][i] = a[rt << 1][i] + a[rt << 1 | 1][i];
        if(a[rt][i] >= mod) a[rt][i] -= mod;
    }
}

inline void push(int rt) {
    lazy[rt] %= 48;
    if(lazy[rt]) {
        lazy[rt << 1] += lazy[rt];
        lazy[rt << 1 | 1] += lazy[rt];
        rotate(a[rt << 1], a[rt << 1] + lazy[rt], a[rt << 1] + 48);
        rotate(a[rt << 1 | 1], a[rt << 1 | 1] + lazy[rt], a[rt << 1 | 1] + 48);
        lazy[rt] = 0;
    }
}

void build(int l, int r, int rt) {
    if(l == r) {
        scanf("%d", &a[rt][0]);
        a[rt][0] %= mod;
        for(int i = 1; i < 48; i++)
            a[rt][i] = 1LL * a[rt][i - 1] * a[rt][i - 1] % mod * a[rt][i - 1] % mod;
        return;
    }
    int mid = l + r >> 1;
    build(lson); build(rson);
    pull(rt);
}

void update(int L, int R, int l, int r, int rt) {
    if(l >= L && r <= R) {
        lazy[rt]++;
        rotate(a[rt], a[rt] + 1, a[rt] + 48);
        return;
    }
    push(rt);
    int mid = l + r >> 1;
    if(L <= mid) update(L, R, lson);
    if(R > mid)  update(L, R, rson);
    pull(rt);
}

int query(int L, int R, int l, int r, int rt) {
    if(l >= L && r <= R) return a[rt][0];
    push(rt);
    int mid = l + r >> 1;
    if(R <= mid) return query(L, R, lson);
    else if(L > mid) return query(L, R, rson);
    else return (query(L, R, lson) + query(L, R, rson)) % mod;
}

int main() {
    scanf("%d", &n);
    build(1, n, 1);
    scanf("%d", &q);
    while(q--) {
        int t, L, R;
        scanf("%d%d%d", &t, &L, &R);
        if(t == 1) {
            printf("%d
", query(L, R, 1, n, 1));
        } else {
            update(L, R, 1, n, 1);
        }
    }
    return 0;
 }

/*
*/
原文地址:https://www.cnblogs.com/CJLHY/p/10616204.html