题意

题目链接

Sol

把式子拆开，就是求这个东西

那么设\(f(x) = \sum_{i = 1}^n i^k\)，这是个经典的\(k + 1\)多项式，直接差值

式子就可以化成

设\(g(x) = \sum_{i = 1}^n f(x)\)

对\(g\)差分之后实际上也就得到了\(f(x)\)，根据多项式的定义，\(g(x)\)是个\(k+2\)次多项式。

同理我们要求的就是个\(k+3\)次多项式

直接暴力插值就行了

时间复杂度：\(O(Tk^3)\)

#include<bits/stdc++.h>

#define int long long

using namespace std;

const int mod = 1234567891, MAXN = 127;

inline int read() {

    char c = getchar(); int x = 0, f = 1;

    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}

    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();

    return x * f;

}

int T, K, a, N, d, f[MAXN], g[MAXN], x[MAXN];

int add(int x, int y) {

    if(x + y < 0) return x + y + mod;

    return x + y >= mod ? x + y - mod : x + y;

}

int add2(int &x, int y) {

    if(x + y < 0) x = (x + y + mod);

    else x = (x + y >= mod ? x + y - mod : x + y);

}

int mul(int x, int y) {

    return 1ll * x * y % mod;

}

int fp(int a, int p) {

    int base = 1;

    while(p) {

        if(p & 1) base = mul(base, a);

        a = mul(a, a); p >>= 1;

    }

    return base;

}

int Large(int *a, int k, int N) {

    for(int i = 0; i <= k; i++) x[i] = i;

    int ans = 0;

    for(int i = 0; i <= k; i++) {

        int up = a[i], down = 1;

        for(int j = 0; j <= k; j++) {

            if(i == j) continue;

            up = mul(up, add(N, -x[j]));

            down = mul(down, add(x[i], -x[j]));

        }

        add2(ans, mul(up, fp(down, mod - 2)));

    }

    return ans;

}

signed main() {

#ifndef ONLINE_JUDGE

    //freopen("a.in", "r", stdin);freopen("a.out", "w", stdout);

#endif

    T = read();

    while(T--) {

        K = read(), a = read(), N = read(), d = read();

        memset(f, 0, sizeof(f)); memset(g, 0, sizeof(g));

        /*

		for(int i = 1; i <= K + 4; i++) f[i] = add(f[i - 1], fp(i, K));

        for(int i = 1; i <= K + 4; i++) g[i] = add(g[i - 1], Large(f, K + 4, a + i * d));//ֱ直接这样写是错的

       	for(int i = 1; i <= K + 4; i++) f[i] = add(f[i - 1], Large(g, K + 4, i));

        printf("%d\n", Large(g, K + 4, N));

    	*/

    	for(int i = 1; i <= K + 4; i++) f[i] = add(f[i - 1], fp(i, K));

    	for(int i = 1; i <= K + 4; i++) f[i] = add(f[i], f[i - 1]);

    	for(int i = 0; i <= K + 4; i++) g[i] = add(i > 0 ? g[i - 1] : 0, Large(f, K + 4, add(a, mul(i, d))));

		printf("%lld\n", Large(g, K + 4, N));

	}

    return 0;

}

/*

5

123 123456789 456879 132

1 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1

*/