CF407C Curious Array（n阶差分）

题目

给出$n$个数，有$m$个操作，每个操作是将$[L,R]$之间的数加上$C(j-L+k,k)$，$L<=j<=R$最后输出这$n$个数的值。

题解

• 虽然我考场上推出了式子但是我还是打的n²暴力（暴力选手qwq
• n阶差分可能有点难以理解可以先看一下$三步必杀（二阶差分）$$三步必杀题解$
• 我们要加的序列
  $$\left(\genfrac{}{}{0ex}{}{k}{k}\right),\left(\genfrac{}{}{0ex}{}{k+1}{k}\right)\cdots \left(\genfrac{}{}{0ex}{}{r-l+k}{k}\right)$$
• 又可以写成
  $$\left(\genfrac{}{}{0ex}{}{k}{0}\right),\left(\genfrac{}{}{0ex}{}{k+1}{1}\right)\cdots \left(\genfrac{}{}{0ex}{}{r-l+k}{r-l}\right)$$
• 结合公式
  $$\left(\genfrac{}{}{0ex}{}{n}{m}\right)=\left(\genfrac{}{}{0ex}{}{n-1}{m}\right)+\left(\genfrac{}{}{0ex}{}{n-1}{m-1}\right)$$
• 该序列的一阶差分序列为
  $$\left(\genfrac{}{}{0ex}{}{k-1}{0}\right),\left(\genfrac{}{}{0ex}{}{k}{1}\right)\cdots \left(\genfrac{}{}{0ex}{}{r-l+k-1}{r-l}\right)\iff \left(\genfrac{}{}{0ex}{}{k-1}{k-1}\right),\left(\genfrac{}{}{0ex}{}{k}{k-1}\right)\cdots \left(\genfrac{}{}{0ex}{}{r-l+k-1}{k-1}\right)$$
• 二阶差分序列
  $$\left(\genfrac{}{}{0ex}{}{k-2}{0}\right),\left(\genfrac{}{}{0ex}{}{k-1}{1}\right)\cdots \left(\genfrac{}{}{0ex}{}{r-l+k-2}{r-l}\right)\iff \left(\genfrac{}{}{0ex}{}{k-2}{k-2}\right),\left(\genfrac{}{}{0ex}{}{k-1}{k-2}\right)\cdots \left(\genfrac{}{}{0ex}{}{r-l+k-2}{k-2}\right)$$
• $j$阶差分序列
  $$\left(\genfrac{}{}{0ex}{}{k-j}{0}\right),\left(\genfrac{}{}{0ex}{}{k-j+1}{1}\right),\cdots ,\left(\genfrac{}{}{0ex}{}{r-l+k-j}{r-l}\right)\iff \left(\genfrac{}{}{0ex}{}{k-j}{k-j}\right),\left(\genfrac{}{}{0ex}{}{k-j+1}{k-j}\right)\cdots \left(\genfrac{}{}{0ex}{}{r-l+k-j}{k-j}\right)$$
• 我们可以发现这样的组合数序列在进行$k+1$次差分后全部变为$0$，我们可以在$k+1$次差分序列的$l$位置加上$1$，在$r+1$位置减去前面的前缀和即可
• 而$j$阶差分的前$k$位的和等于$j-1$阶差分序列的第$k$个数（手推一下吧
• 所以第$j$阶序列的前缀和
  $$\sum _{j=0}^k\left(\genfrac{}{}{0ex}{}{r-l+k-j}{r-l}\right)=\left(\genfrac{}{}{0ex}{}{r-l+k-j+1}{k-j+1}\right)$$
• 每次消除前缀和对后面的影响即可

$code$

#include 
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#include 
using namespace std; 
const int maxn = 100000 + 1000;
const int maxm = 100 + 10;
const int mod = 1000000007;
typedef long long LL; 

template <class T> 
inline void read(T &s) {
	s = 0; 
	T w = 1, ch = getchar(); 
	while (!isdigit(ch)) { if (ch == '-') w = -1; ch = getchar(); }
	while (isdigit(ch)) { s = (s << 1) + (s << 3) + (ch ^ 48); ch = getchar(); }
	s *= w;  
}

int n, m; 
LL a[maxn]; 
LL c[maxn][maxm], ans[maxn][maxm]; 

void init() {
	c[0][0] = 1; 
	for (int i = 1; i < maxn; ++i) {
		c[i][0] = 1; 
		for (int j = 1; j <= i && j < maxm; ++j) {
			c[i][j] = (c[i-1][j-1] + c[i-1][j]) % mod; 
		}
	}
}

int main() {
	freopen("data.in", "r", stdin); 
	freopen("bf.out", "w", stdout); 

	init(); 
	scanf("%d%d", &n, &m); 
	for (int i = 1; i <= n; ++i) 
		scanf("%d", &a[i]);  
	int l, r, k; 
	while (m--) { 
		scanf("%d %d %d", &l, &r, &k); 
		for (int i = 0; i <= k; ++i) 
			ans[l][i] = (ans[l][i] + c[k][k-i]) % mod; 
		for (int i = 0; i <= k; ++i)
			ans[r+1][i] = (ans[r+1][i] - c[r-l+k+1][k-i]) % mod; 
	}
	for (int i = 1; i < n; ++i) {
		for (int j = 101; j >= 1; --j) {
			ans[i+1][j-1] = (ans[i+1][j-1] + ans[i][j] + ans[i][j-1]) % mod; 
		}
	}
	for (int i = 1; i <= n; ++i) {
		printf("%lld ", ((ans[i][0] + a[i]) % mod + mod) % mod); 
	}
	return 0; 
}