解题报告:可以说是一个纯数学题,要用到二元一次和二元二次解方程,我们假设[a,b]这个区间的所有的数的和是N,由此,我们可以得到以下公式: (b-a+1)*(a+b) / 2 = N;很显然,这是一个二元一次方程,如果可以解出这个方程里的a,b就可以了,N是已知的,那么可以怎么一个公式怎么解二元方程呢?这里有两种方法可以选择,第一种是可以枚举a = 1,2,3.....,一直枚举到N/2,a是不可能大于N/2,这个你可以自己证明一下,但是这样做的缺点就是数据量还是太大了,N/2任然达到了10的八…

The sum problem Problem Description Given a sequence 1,2,3,......N, your job is to calculate all the possible sub-sequences that the sum of the sub-sequence is M.   Input Input contains multiple test cases. each case contains two integers N, M( 1 <=…

一个数学问题:copy了别人的博客 #include<cstdio> #include<cstdlib> #include<cmath> int main() { int n,m,i,j; while(~scanf("%d%d",&n,&m)) { for( j=(int)sqrt(2*m); j>=1;j--) { i= (2*m/j-j+1)/2; if((i*2-1+j)*j == 2*m && i+j-1…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2058 问题描述 给定一个序列1,2,3,...... N,你的工作是计算所有可能的子序列,其子序列的总和为M. 输入 输入包含多个测试用例. 每个情况包含两个整数N,M(1 <= N,M <= 1000000000).输入以N = M = 0结束. 输出  对于每个测试用例,打印其总和为M的所有可能的子序列.格式显示在下面的示例中.在每个测试用例后打印一个空行. 示例输入 20 10 50 30…

Problem Description Given a sequence 1,2,3,......N, your job is to calculate all the possible sub-sequences that the sum of the sub-sequence is M.   Input Input contains multiple test cases. each case contains two integers N, M( 1 <= N, M <= 1000000…

传送门 Description Given a sequence 1,2,3,......N, your job is to calculate all the possible sub-sequences that the sum of the sub-sequence is M. Input Input contains multiple test cases. each case contains two integers N, M( 1 <= N, M <= 1000000000).i…

题目: Given a sequence 1,2,3,......N, your job is to calculate all the possible sub-sequences that the sum of the sub-sequence is M. 思路: 刚开始做写了一个尺取法,脑抽的一批.(这几天心情真的是颓的很,连带脑瓜也不好用了,抓紧调整~~~~) 其实可以根据等差数列求和公式求出这个子序列的长度的一个大致的范围为k=sqrt(2*m),然后根据m和k求出首项a1,然后将a1…

题意:给定一个N和M,N表示从1到N的连续序列,让你求在1到N这个序列中连续子序列的和为M的子序列区间. 析:很明显最直接的方法就是暴力,可是不幸的是,由于N,M太大了,肯定会TLE的.所以我们就想能不能优化一下,找一个范围.想到这是一个连续的序列而且是从1开始的,这不就是一个等差数列么,公差是1罢了.由求和公式得Sn = (a1+an) * n / 2;所以说n最大就是sqrt(M*2)(想一想为什么),因为a1+an 一定是大于n的.如果我们取区间的和,那么Sn = (ai+aj) * (j…

题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=2058 思路: 这题的n,m都很大,很显然直接暴力,会超时,那就不能全部都找了,利用等差数列求和公式, (1)sn=n*(a1+an)/2: 即可代入公式,(2)m=(e-s+1)*(s+e)/2                         注释*******//s代表起点,e代表终点. 则由(2)推出,(3)e=(int)(2*m+s*s-s),根据e点找s点,代入(1)成立则输出[s,e]; …

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4961 Problem Description Number theory is interesting, while this problem is boring. Here is the problem. Given an integer sequence a1, a2, -, an, let S(i) = {j|1<=j<i, and aj is a multiple of ai}. If S…

Problem Description Given a sequence 1,2,3,--N, your job is to calculate all the possible sub-sequences that the sum of the sub-sequence is M. Input Input contains multiple test cases. each case contains two integers N, M( 1 <= N, M <= 1000000000).i…

题意:给一个n*m的矩阵,每个格子中有一个数字,每个格子仅可以走一次,问从(1,1)走到(n,m) 的路径点权之和. 思路: 想了挺久,就是有个问题不能短时间证明,所以不敢下手. 显然只要n和m其中一个是奇数,逐行/列绕就可以到达终点,可是恰好都是偶数呢?由于绕不到,那至少得舍弃1个,但是弃哪个比较好?况且有些格子是弃不了的(画4*4的模拟就知道了). 通过画图可以知道(自己绕!),行号+列号为奇数的格子都是可以舍弃的,而且可以保证其他所有格子都能走一遍到终点(无论是从行/列为单位来绕,这个图都…

The sum problem Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 31453    Accepted Submission(s): 9414 Problem Description Given a sequence 1,2,3,......N, your job is to calculate all the possib…

Sum Problem Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 338086    Accepted Submission(s): 85117 Problem Description Hey, welcome to HDOJ(Hangzhou Dianzi University Online Judge). In this pro…

HDOJ(HDU).1016 Prime Ring Problem (DFS) [从零开始DFS(3)] 从零开始DFS HDOJ.1342 Lotto [从零开始DFS(0)] - DFS思想与框架/双重DFS HDOJ.1010 Tempter of the Bone [从零开始DFS(1)] -DFS四向搜索/奇偶剪枝 HDOJ(HDU).1015 Safecracker [从零开始DFS(2)] -DFS四向搜索变种 HDOJ(HDU).1016 Prime Ring Problem (…

A - Max Sum Plus Plus Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1024 Appoint description: Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a bra…

/* hdu 1402 A * B Problem Plus FFT 这是我的第二道FFT的题 第一题是完全照着别人的代码敲出来的,也不明白是什么意思 这个代码是在前一题的基础上改的 做完这个题,我才有点儿感觉,原来FFT在这里就是加速大整数乘法而已 像前一题,也是一个大整数乘法,然后去掉一些非法的情况 */ #pragma warning(disable : 4786) #pragma comment(linker, "/STACK:102400000,102400000") #in…

题目链接:hdu 2993 MAX Average Problem 题意: 给一个长度为 n 的序列,找出长度 >= k 的平均值最大的连续子序列. 题解: 这题是论文的原题,请参照2004集训队论文<周源--浅谈数形结合思想在信息学竞赛中的应用> 这题输入有点大,要加读入优化才能过. #include<bits/stdc++.h> #define F(i,a,b) for(int i=a;i<=b;++i) using namespace std; int tot;…

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…

hdu 5618 Jam's problem again #include <bits/stdc++.h> #define MAXN 100010 using namespace std; int n,k,T,xx; int ans[MAXN],c[MAXN],f[MAXN]; struct Node{ int x,y,z,id; }a[100010],b[100010]; inline int read(){ char ch; bool f=false; int res=0; while (…

HDU 1024 Max Sum Plus Plus (动态规划) Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem. Given…

Max Sum Plus Plus Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1024 Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to mor…

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1024 Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 35988    Accepted Submission(s): 12807 Problem Description Now I think you ha…

数字的反转: 就是将数字倒着存下来而已.(*^__^*) 嘻嘻…… 大致思路:将数字一位一位取出来,存在一个数组里面,然后再将其变成数字,输出. 详见代码. while (a) //将每位数字取出来,取完为止 { num1[i]=a%; //将每一个各位取出存在数组里面,实现了将数字反转 i++; //数组的变化 a/=; } 趁热打铁 例题:hdu 4554 叛逆的小明 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4554 叛逆的小明 Time…

[HDU 5293]Tree chain problem(树形dp+树链剖分) 题面 在一棵树中,给出若干条链和链的权值,求选取不相交的链使得权值和最大. 分析 考虑树形dp,dp[x]表示以x为子树的最大权值和(选的链都在i的子树中) 设sum[x]表示x的儿子的dp值和,即\(\sum _{y \in \mathrm{son}(x)} dp[y]\) 1.不选两端点lca为x的链,dp[x]=sum[x] 2.选两端点lca为x的链,则dp[x]=max{链的权值+链上节点的所有子节点dp的…

I found summary of k Sum problem and solutions in leetcode on the Internet. http://www.sigmainfy.com/blog/summary-of-ksum-problems.html…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 25639    Accepted Submission(s): 8884 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem.…

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https://en.wikipedia.org/wiki/Subset_sum_problem In computer science, the subset sum problem is an important problem in complexity theory and cryptography. The problem is this: given a set (or multiset) of integers, is there a non-empty subset whose…

HDU 1024 题目大意:给定m和n以及n个数,求n个数的m个连续子系列的最大值,要求子序列不想交. 解题思路:<1>动态规划,定义状态dp[i][j]表示序列前j个数的i段子序列的值,其中第i个子序列包括a[j], 则max(dp[m][k]),m<=k<=n 即为所求的结果 <2>初始状态: dp[i][0] = 0, dp[0][j] = 0; <3>状态转移: 决策:a[j]自己成为一个子段,还是接在前面一个子段的后面 方程: a[j]直接接在前面…