CSU-2173 Use FFT

CSU-2173 Use FFT

Description

Bobo computes the product P(x)⋅Q(x)=(c_0 + c_1x + … + c_{n+m}x^{n + m}) for two polynomials P(x)=(a_0 + a_1x + … + a_nx^n) and Q(x)=(b_0 + b_1x + … + b_mx^m). Find $ (c_L + c_{L + 1} + … + c_R) $ modulo ($10^9 $ + 7) for given L and R.

• 1 ≤ n, m ≤ 5 × (10^5)
• 0 ≤ L ≤ R ≤ n + m
• 0 ≤ (a_i, b_i) ≤ (10^9)
• Both the sum of n and the sum of m do not exceed (10^6).

Input

The input consists of several test cases and is terminated by end-of-file.

The first line of each test case contains four integers n, m, L, R.

The second line contains (n + 1) integers (a_0, a_1, …, a_n).

The third line contains (m + 1) integers (b_0, b_1, …, b_m).

Output

For each test case, print an integer which denotes the reuslt.

Sample Input

1 1 0 2
1 2
3 4
1 1 1 2
1 2
3 4
2 3 0 5
1 2 999999999
1 2 3 1000000000

Sample Output

21
18
5

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题解

这题标题是Use FFT所以当然是用FFT做了(滑稽)

这题其实是个数学题+找规律题，借用一张图片

所以我们对b求前缀和，用a去乘，注意细节就好了

#include<bits/stdc++.h>
#define maxn 500050
#define p 1000000007
using namespace std;
typedef long long ll;
ll a[maxn], b[maxn];
ll pre[maxn * 2];
int main() {
	int n, m, l, r;
	while (scanf("%d%d%d%d", &n, &m, &l, &r) != EOF) {
		for (int i = 1; i <= n + 1; i++) {
			scanf("%lld", &a[i]);
		}
		for (int i = 1; i <= m + 1; i++) {
			scanf("%lld", &b[i]);
			pre[i] = (pre[i - 1] + b[i]) % p;
		}
		for (int i = m + 2; i <= r + 1; i++) {
			pre[i] = pre[i - 1];
		}
		ll ans = 0;
		for (int i = 1; i <= n + 1; i++) {
			ans = (ans + a[i] * (pre[r + 1] - pre[l] + p) % p) % p;
			if (l > 0) l--;
			if (r >= 0) r--;
		}
		printf("%lld
", (ans + p) % p);
	}
	return 0;
}