Power Sum Time Limit: 20000/10000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others) SubmitStatus Problem Description 给出n,m,p,求 (1^m + 2^m + 3^m + 4^m + ... + n^m) % p Input 第一行一个数T( <= 10),表示数据总数 然后每行给出3个数n,m,p(1 <= n <= m <= 10^18, 1…

先上题目: Power Sum Time Limit: 20000/10000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others) SubmitStatus Problem Description 给出n,m,p,求 (1^m + 2^m + 3^m + 4^m + ... + n^m) % p Input 第一行一个数T( <= 10),表示数据总数 然后每行给出3个数n,m,p(1 <= n <= m <= 10…

FFT #include<iostream> #include<cstring> #include<cstdlib> #include<cstdio> #include<cmath> #include<algorithm> #define maxn 1000005 using namespace std; inline int read() { ,f=;char ch=getchar(); ; +ch-'; return x*f; }…

Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number. An example is the root-to-leaf path 1->2->3 which represents the number 123. Find the total sum of all root-to-leaf numbers. For example, 1 / \ 2 3…

题目链接:http://lightoj.com/volume_showproblem.php?problem=1278 题意:给你一个数n(n<=10^14),然后问n能用几个连续的数表示; 例如: 15 = 7+8 = 4+5+6 = 1+2+3+4+5,所以15对应的答案是3,有三种; 我们现在相当于已知等差数列的和sum = n, 另首项为a1,共有m项,那么am = a1+m-1: sum = m*(a1+a1+m-1)/2  -----> a1 = sum/m - (m-1)/2 a…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 29942    Accepted Submission(s): 10516 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…

Given a binary tree containing digits from0-9only, each root-to-leaf path could represent a number. An example is the root-to-leaf path1->2->3which represents the number123. Find the total sum of all root-to-leaf numbers. For example, 1 / \ 2 3 The…

Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number. An example is the root-to-leaf path 1->2->3 which represents the number 123. Find the total sum of all root-to-leaf numbers. Note: A leaf is a node…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1024 题意就是有n个数分成m段,求最大的和: dp[i][j]表示把 j 个数分成 i 段,选择第 j 个数的结果,而并不是当前的最优解, 那么要考虑的是第i个数的数是自己成一段还是和前面的成一段 所以dp[i][j]=max(dp[i][j-1]+a[j], Max+a[j]); 其中Max为前 j-1 个数字分成 i-1 段中的最大值: 由于题中n为100w,m不知道是多少,所以开二维数组可能不…

给定一个只包含 0-9 数字的二叉树,每个根到叶的路径可以代表一个数字.例如,从根到叶路径 1->2->3则代表数字 123.查找所有根到叶数字的总和.例如,    1   / \  2   3根到叶子路径 1->2 表示数字 12.根到叶子路径 1->3 表示数字 13.返回总和 = 12 + 13 = 25.详见:https://leetcode.com/problems/sum-root-to-leaf-numbers/description/ Java实现: /** * D…