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  • 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
    ", 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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  • 原文地址:https://www.cnblogs.com/CJLHY/p/10885059.html
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