BZOJ3453: tyvj 1858 XLkxc(拉格朗日插值)

题意

题目链接

Sol

把式子拆开，就是求这个东西

[sum_{i = 0} ^n sum_{j = 1}^{a + id} sum_{x =1}^j x^k pmod P ]

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

[sum_{i = 0} ^n sum_{j = 1}^{a + id} f(j) pmod P ]

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
", 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
", 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
*/