考试时候遇到这种题只会找规律 You've got an array a, consisting of n integers. The array elements are indexed from 1 to n. Let's determine a two step operation like that: First we build by the array a an array s of partial sums, consisting of n elements. Element nu…

开始想的是O(n2logk)的算法但是显然会tle.看了解题报告然后就打表找起规律来.嘛是组合数嘛.时间复杂度是O(nlogn+n2)的 #include<cstdio> #include<cstring> #include<cctype> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(in…

Non-negative Partial Sums Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2622    Accepted Submission(s): 860 Problem Description You are given a sequence of n numbers a0,..., an-1. A cyclic shi…

Non-negative Partial Sums Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1420    Accepted Submission(s): 544 Problem Description You are given a sequence of n numbers a0,..., an-1. A cyclic shi…

Description Given a series of n numbers a1, a2, ..., an, the partial sum of the numbers is defined as the sum of ai, ai+1, ..., aj. You are supposed to calculate how many partial sums of a given series of numbers could be divided evenly by a given nu…

Partial Sums 题解: 一个数列多次前缀和之后, 对于第i个数来说他的答案就是 ; i <= n; ++i){ ; j <= i; ++j){ b[i] = (b[i] + 1ll * a[j] * C(k-+j-i,j-i)) % mod; } } 唯一注意的就是这个k会到1e9. 观察可能,其实我们最多也就用了n个组合数, 并且这个C(n, m) 的 m 足够小. 所以我们可以根据定义先把这几个组合数先预处理出来. 代码: #include<bits/stdc++.h>…

1161 Partial Sums  题目来源: CodeForces 基准时间限制:2 秒 空间限制:131072 KB 分值: 80 难度:5级算法题  收藏  取消关注 给出一个数组A,经过一次处理,生成一个数组S,数组S中的每个值相当于数组A的累加,比如:A = {1 3 5 6} => S = {1 4 9 15}.如果对生成的数组S再进行一次累加操作,{1 4 9 15} => {1 5 14 29},现在给出数组A,问进行K次操作后的结果.(每次累加后的结果 mod 10^9 +…

ACM思维题训练集合 You've got an array a, consisting of n integers. The array elements are indexed from 1 to n. Let's determine a two step operation like that: First we build by the array a an array s of partial sums, consisting of n elements. Element number…

题面 传送门 题解 orz zzk 考虑这东西的组合意义 (图片来自zzk) \(a_i\)这个元素对\(k\)阶前缀和的第\(j\)个元素\(s_{k,j}\)的贡献就等于从\((0,i)\)走到\((j,k)\)的方案数(最开始的一次必须往下走,所以实际上是从\((1,i)\)走到\((j,k)\)的方案数) 那么\(s_{j,k}=\sum_{i=1}^ja_i{j-i+k-1\choose j-i}\) 直接\(O(n^2)\)递推就行了 //minamoto #include<bits…

题意: 给定一个由n个整数组成的整数序列,可以滚动,滚动的意思就是前面k个数放到序列末尾去.问有几种滚动方法使得前面任意个数的和>=0. 思路: 先根据原来的数列求sum数组,找到最低点,然后再以最低点为始点,求解题目答案,(每求解一始点i,符合要求的条件为:sum[i]>=minx,[minx是i<x<=n中的最小值],之所以不用考虑前面的,就是因为我们的预处理是的所有的x<i的sum[x]的值满足sum[x]>=sum[i]) 代码如下: #include <…