[CF622F]The Sum of the k-th Powers

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[题解] CF622F The Sum of the k-th Powers

CF622F The Sum of the k-th Powers 题意:给$n$和$k$,让你求$\sum\limits_{i = 1} ^ n i^k \ mod \ 10^9 + 7$.$(1 \le n \le 10^9,0 \le k \le 10^6)$ 好,我们先不看题,来补一些数学. 考虑这样一个序列 \[ h_0,h_1,\dots,h_n,\dots \] 我们定义它的一个差分序列(一阶) \[ \Delta h_0, \Delta h_1, \dots ,…

解题：CF622F The Sum of the k-th Powers

题面 TJOI2018出CF原题弱化版是不是有点太过分了?对,就是 TJOI2018 教科书般的亵渎然而我这个问题只会那个题的范围的m^3做法回忆一下1到n求和是二次的,平方求和公式是三次的,立方求和公式是四次的,那m次方求和公式一定是个m+1次多项式直接扔m+2个值进去把它插出来,因为是连续的可以做到线性(不算逆元) #include<cstdio> ,mod=1e9+; int n,k,ans,sum,fac[N],pre[N],suf[N]; int Qpow(int x,int…

[Swift]LeetCode862. 和至少为 K 的最短子数组 | Shortest Subarray with Sum at Least K

Return the length of the shortest, non-empty, contiguous subarray of A with sum at least K. If there is no non-empty subarray with sum at least K, return -1. Example 1: Input: A = [1], K = 1 Output: 1 Example 2: Input: A = [1,2], K = 4 Output: -1 Exa…

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leetcode 862 shorest subarray with sum at least K

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862. Shortest Subarray with Sum at Least K

Return the length of the shortest, non-empty, contiguous subarray of A with sum at least K. If there is no non-empty subarray with sum at least K, return -1. Example 1: Input: A = [1], K = 1 Output: 1 Example 2: Input: A = [1,2], K = 4 Output: -1 Exa…

[LeetCode] 862. Shortest Subarray with Sum at Least K 和至少为K的最短子数组

Return the length of the shortest, non-empty, contiguous subarray of A with sum at least K. If there is no non-empty subarray with sum at least K, return -1. Example 1: Input: A = [1], K = 1 Output: 1 Example 2: Input: A = [1,2], K = 4 Output: -1 Exa…

【LeetCode】1099. Two Sum Less Than K 解题报告(C++)

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[CF622F]The Sum of the k-th Powers

题目大意:给你$n,k(n\leqslant10^9,k\leqslant10^6)$,求:$$\sum\limits_{i=1}^ni^k\pmod{10^9+7}$$ 题解:可以猜测是一个$k+1$次的多项式,可以求出$x$在$0,1,2,3,\dots,k+1$时的值,设为$s_0,s_1,\dots,s_{k+1}$,根据拉格朗日插值公式: $$\begin{align*}f_n&=\sum\limits_{i=0}^{k+1}y_i\prod\limits_{j=0,j\not=i}^…

CF622F The Sum of the k-th Powers（拉格朗日插值）

题意给出 $n,k$ , $n\le10^9,k\le10^6$ ,求 $\sum_{i=1}^n i^k(mod\;10^9+7)$ 题解自然数幂次和,是一个$k+1$次多项式,那么算出$k+2$个值然后差值就行了 //minamoto #include<bits/stdc++.h> #define R register #define fp(i,a,b) for(R int i=a,I=b+1;i<I;++i) #define fd(i,a,b) for(R…

[LeetCode] Partition to K Equal Sum Subsets 分割K个等和的子集

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LeetCode 1099. Two Sum Less Than K

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Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k.Example 1:Input:nums = [1,1,1], k = 2Output: 2Note:The length of the array is in range [1, 20,000].The range of numbers in th…