题目链接: http://codeforces.com/problemset/problem/466/C 题意: 给一个长度为n的数组,将其分成连续的三段使三段的和相等.求有几种这种组合 分析: 从头扫到尾.将全部的前缀和为(sum/3)的点统计起来.然后再从尾開始统计.找到统计全部后缀和为(sum/3)的节点 然后这样的方案的数为 这个点之前全部前缀和为sum/3的个数 代码例如以下: #include <iostream> #include <cstring> #include…

C. Number of Ways time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You've got array a[1], a[2], ..., a[n], consisting of n integers. Count the number of ways to split all the elements of th…

You've got array a[1], a[2], ..., a[n], consisting of n integers. Count the number of ways to split all the elements of the array into three contiguous parts so that the sum of elements in each part is the same. More formally, you need to find the nu…

题目链接:5274. 停在原地的方案数 You have a pointer at index 0 in an array of size arrLen. At each step, you can move 1 position to the left, 1 position to the right in the array or stay in the same place (The pointer should not be placed outside the array at any…

题目如下: You have a pointer at index 0 in an array of size arrLen. At each step, you can move 1 position to the left, 1 position to the right in the array or stay in the same place  (The pointer should not be placed outside the array at any time). Given…

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #define N 500005 using namespace std; typedef long long LL; LL prefix[N], suffix[N], num[N]; LL cntSuf[N]; int main(){ int n; scanf("%d", &n);…

题目链接:http://codeforces.com/problemset/problem/466/C 题目意思:给出一个 n 个数的序列你,问通过将序列分成三段,使得每段的和都相等的分法有多少种. 这个是详细的题解: http://codeforces.com/blog/entry/13758 代码也是按这个思路来做的,可怜我的做法改了 N 次总是得不到正确的结果--泪 #include <iostream> #include <cstdio> #include <cstd…

https://codeforces.com/problemset/problem/466/C 要把数据分为均等的非空的三组,那么每次确定第二个分割点的时候把(除此之外的)第一个分割点的数目加上就可以了.记得最后给第三组留至少一个. #include<bits/stdc++.h> using namespace std; #define ll long long int n; ]; int main(){ scanf("%d",&n); ;i<n;i++) s…

[题目链接] https://codeforces.com/contest/466/problem/C [算法] 维护序列前缀和 , 枚举中间一段即可 , 详见代码 时间复杂度 : O(N) [代码] #include<bits/stdc++.h> using namespace std; ; int a[MAXN]; long long sum[MAXN]; template <typename T> inline void read(T &x) { T f = ; x…

[链接] 我是链接,点我呀:) [题意] 让你把数组分成3个连续的部分 每个部分的和要一样 问你有多少种分法 [题解] 先处理出来num[i] 表示i..n这里面有多少个j 满足aft[j] = aft[i]/2 这aft[i]=a[j]+a[j+1]..+a[n] 然后for从1..n 看看pre[i]*2=aft[i+1]是否成立. 如果成立的话那么答案就加上num[i+1] [代码] import java.io.*; import java.util.*; public class Ma…