Codeforces 266E More Queries to Array... 线段树 + 二项式展开

把式子二项式展开之后， 会发现是需要维护a[ i ],  i * a[ i ] ....   i ^ 5 * a[ i ]， 的区间和， 然后用线段树维护。

#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 = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1000000007;
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;}

int power(int a, int b) {
    int ans = 1;
    while(b) {
        if(b & 1) ans = 1LL * ans * a % mod;
        a = 1LL * a * a % mod; b >>= 1;
    }
    return ans;
}

int n, m;
int a[N];
int prefix[6][N];
int C[6][6];
char op[3];

struct SegmentTree {
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
    int sum[N << 2], lazy[N << 2];
    inline void pull(int rt) {
        sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
        if(sum[rt] >= mod) sum[rt] -= mod;
    }
    inline void push(int Pow, int rt, int l, int r) {
        if(~lazy[rt]) {
            int mid = l + r >> 1;
            sum[rt << 1] = 1LL * lazy[rt] * (prefix[Pow][mid] - prefix[Pow][l - 1] + mod) % mod;
            sum[rt << 1 | 1] = 1LL * lazy[rt] * (prefix[Pow][r] - prefix[Pow][mid] + mod) % mod;
            lazy[rt << 1] = lazy[rt];
            lazy[rt << 1 | 1] = lazy[rt];
            lazy[rt] = -1;
        }
    }
    void build(int Pow, int l, int r,  int rt) {
        lazy[rt] = -1;
        if(l == r) {
            sum[rt] = 1LL * a[l] * power(l, Pow) % mod;
            return;
        }
        int mid = l + r >> 1;
        build(Pow, lson); build(Pow, rson);
        pull(rt);
    }
    void update(int L, int R, int val, int Pow, int l, int r, int rt) {
        if(R < l || r < L || R < L) return;
        if(L <= l && r <= R) {
            sum[rt] = 1LL * val * (prefix[Pow][r] - prefix[Pow][l - 1] + mod) % mod;
            lazy[rt] = val;
            return;
        }
        push(Pow, rt, l, r);
        int mid = l + r >> 1;
        update(L, R, val, Pow, lson);
        update(L, R, val, Pow, rson);
        pull(rt);
    }
    LL query(int Pow, int L, int R, int l, int r, int rt) {
        if(R < l || r < L || R < L) return 0;
        if(L <= l && r <= R) return sum[rt];
        push(Pow, rt, l, r);
        int mid = l + r >> 1;
        return (query(Pow, L, R, lson) + query(Pow, L, R, rson)) % mod;
    }
} Tree[6];

int main() {
    for(int p = 0; p < 6; p++)
        for(int i = 1; i < N; i++)
            prefix[p][i] = (prefix[p][i - 1] + power(i, p)) % mod;
    for(int i = 0; i < 6; i++)
        for(int j = C[i][0] = 1; j < 6; j++)
            C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for(int i = 0; i < 6; i++) Tree[i].build(i, 1, n, 1);
    int l, r, x, k;
    while(m--) {
        scanf("%s", op);
        if(*op == '?') {
            scanf("%d%d%d", &l, &r, &k);
            int c = (1 - l + mod) % mod;
            int ans = 0;
            for(int z = 0; z <= k; z++) {
                int val = Tree[z].query(z, l, r, 1, n, 1);
                add(ans, 1LL * val * C[k][z] % mod * power(c, k - z) % mod);
            }
            printf("%d\n", ans);
        } else {
            scanf("%d%d%d", &l, &r, &x);
            for(int i = 0; i < 6; i++) Tree[i].update(l, r, x, i, 1, n, 1);
        }
    }
    return 0;
}

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