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Sums
poj 3187 Backward Digit
Sums
DescriptionFJandhiscowsenjoyplayingamentalgame.Theywritedownthenumbersfrom1toN(1 #include #include #include usingnamespacestd; intn,sum,dp[20][20]; intmain() { while(cin>>n>>sum) { memset(dp,0,sizeof(
qq_33638791
·
2016-04-20 11:00
Python 练习实例25
sums
=0 n=int(input(u'前n项和:')) forminrange(1,n+1): sum=1 forninrange(1,n+1): ifn<=m: sum*=n
sums
+=sum
mwmoo
·
2016-04-16 14:00
python
实例
poj 3178 Backward Digit
Sums
(全排列)
BackwardDigitSumsTimeLimit: 1000MS MemoryLimit: 65536KTotalSubmissions: 5858 Accepted: 3398DescriptionFJandhiscowsenjoyplayingamentalgame.Theywritedownthenumbersfrom1toN(13124 436 79 16BehindFJ'sba
bmicnj
·
2016-04-14 13:00
CF348C Subset
Sums
思路好题首先,将集合大小大于等于sqrt(n)的集合称为重集合,其余称为轻集合,则重集合个数小于等于sqrt(n)。令f[i][j]表示第i个集合和第j个集合的相同元素个数,这可以在O(n*sqrt(n))内预处理。再令sum[i]表示第i个重集合的元素和,add[i]表示第i个重集合所有元素的增量标记。修改操作,重集合直接修改sum[i]和add[i],轻集合暴力修改,同时用f[i][j]修改重
AaronGZK
·
2016-04-13 00:00
codeforces
CF348C Subset
Sums
思路好题首先,将集合大小大于等于sqrt(n)的集合称为重集合,其余称为轻集合,则重集合个数小于等于sqrt(n)。令f[i][j]表示第i个集合和第j个集合的相同元素个数,这可以在O(n*sqrt(n))内预处理。再令sum[i]表示第i个重集合的元素和,add[i]表示第i个重集合所有元素的增量标记。修改操作,重集合直接修改sum[i]和add[i],轻集合暴力修改,同时用f[i][j]修改重
AaronGZK
·
2016-04-13 00:00
codeforces
LeetCode之16----3
Sums
Closest
题目:Givenanarray S of n integers,findthreeintegersin S suchthatthesumisclosesttoagivennumber,target.Returnthesumofthethreeintegers.Youmayassumethateachinputwouldhaveexactlyonesolution.Forexample,givena
Jung_zhang
·
2016-04-12 19:00
LeetCode
算法
LeetCode之16----3
Sums
Closest
题目:Givenanarray S of n integers,findthreeintegersin S suchthatthesumisclosesttoagivennumber,target.Returnthesumofthethreeintegers.Youmayassumethateachinputwouldhaveexactlyonesolution.Forexample,givena
Jung_zhang
·
2016-04-12 19:00
LeetCode
算法
【POJ】3187 - Backward Digit
Sums
(枚举)
BackwardDigitSumsTimeLimit: 1000MS MemoryLimit: 65536KTotalSubmissions: 5835 Accepted: 3385DescriptionFJandhiscowsenjoyplayingamentalgame.Theywritedownthenumbersfrom1toN(1 #include usingnamespacestd;
wyg1997
·
2016-04-10 12:00
【暑假】[实用数据结构]UVa11997 K Smallest
Sums
UVa11997KSmallestSums 题目: KSmallestSums You'regivenkarrays,eacharrayhaskintegers.Therearekk waystopickexactlyoneelementineacharrayandcalculatethes
hahalidaxin
·
2016-03-30 17:00
lightoj 1125 - Divisible Group
Sums
01背包变形
给定一个n,q以及n个数字,代表有q次询问,每次询问一个d,m,选择m个数字(不可重复选)组成集合是d的倍数有多少种方法q比较小可以先不考虑...对于一个数字选与不选就构成01背包...选择了多少个数字当做背包体积吧...因为要求方案数,所以不能拿余数当做价值了...得另外再开一维当做余数,那么选了确定x个数字后余数是多少也要顺便枚举一下吧...注意,01背包的滚动数组体积那重循环是逆向的,所以选
azx736420641
·
2016-03-30 14:00
dp
01背包
lightoj
HDU 4193 Non-negative Partial
Sums
【单调队列】
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4193题意:给定序列,可以把后面的连续的部分移到最前面来,问多少种移法使得最终得到的序列的前i项和均大于等于0(1≤i≤n)?分析:先将数组前后拼接起来,记录每个位置的前缀和。然后去找长度为n的滑动窗口,在这个窗口内,要保证前i项和大于等于0,也即区间[i,n+i]之间的找到的最小的前缀和要比a[i−1
Yukizzz
·
2016-03-30 11:00
hdoj-2715-Herd
Sums
DescriptionThecowsinfarmerJohn'sherdarenumberedandbrandedwithconsecutiveintegersfrom1toN(1 intmain() { intk,i,j,sum,num; while(scanf("%d",&k)!=EOF) { num=0; for(i=1;ik) break; } } printf("%d\n",num);
Ypopstar
·
2016-03-27 11:00
模拟
POJ 3187 Backward Digit
Sums
(全排列)
BackwardDigitSumsTimeLimit:1000MS MemoryLimit:65536KTotalSubmissions:5727 Accepted:3320DescriptionFJandhiscowsenjoyplayingamentalgame.Theywritedownthenumbersfrom1toN(1 #include #include #include #incl
helloiamclh
·
2016-03-16 08:00
poj 3187 Backward Digit
Sums
(dfs暴搜)
问题描述FJandhiscowsenjoyplayingamentalgame.Theywritedownthenumbersfrom1toN(1 #include usingnamespacestd; intn,sum,c[11],dp[11];intvis[11]; boolflag; intfond_c(inta,intb)//求C(a,b) { intans=1;intm=b; for
qq_31237061
·
2016-03-08 21:00
POJ 3187 Backward Digit
Sums
http://poj.org/problem?id=3187BackwardDigitSumsTimeLimit: 1000MS MemoryLimit: 65536KTotalSubmissions: 5684 Accepted: 3293DescriptionFJandhiscowsenjoyplayingamentalgame.Theywritedownthenumbersfrom1toN(
yuzhiwei1995
·
2016-03-08 20:00
UVa 11997 K Smallest
Sums
优先队列&&打有序表&&归并
UVA-11997KSmallestSumsTimeLimit: 1000MS MemoryLimit: Unknown 64bitIOFormat: %lld&%lluSubmit StatusYou’regivenkarrays,eacharrayhaskintegers.Therearek^kwaystopickexactlyoneelementineacharrayandcalculate
ProLightsfxjh
·
2016-03-05 13:00
数据结构
ACM
优先队列
uva
有序表
UVa 11997 K Smallest
Sums
优先队列&&打有序表&&归并
UVA-11997KSmallestSumsTimeLimit: 1000MS MemoryLimit: Unknown 64bitIOFormat: %lld&%lluSubmit StatusYou’regivenkarrays,eacharrayhaskintegers.Therearek^kwaystopickexactlyoneelementineacharrayandcalculate
ProLightsfxjh
·
2016-03-05 13:00
数据结构
ACM
优先队列
uva
有序表
Backward Digit
Sums
(暴力)
BackwardDigitSumsTimeLimit: 1000MS MemoryLimit: 65536KTotalSubmissions: 5664 Accepted: 3280DescriptionFJandhiscowsenjoyplayingamentalgame.Theywritedownthenumbersfrom1toN(1 #include #include #includ
handsomecui
·
2016-03-05 13:00
Different
Sums
2065.DifferentSumsTimelimit:1.0secondMemorylimit:64MBAlexisaveryseriousmathematicianandhelikestosolveseriousproblems.Forexample,thisproblem.Youaretoconstructanarrayof n integersinwhichtheamountofdiffe
功亏一篑
·
2016-03-04 10:00
POJ 3187 Backward Digit
Sums
POJ3187题目大意如下:一种智力游戏,给出一个数finalSum和自然数N,前者表示通过游戏规则得到的和,后者表示最开始一层可以使用的1~N这几个数N最大可以到10,想到10!规模太大,所以一开始没有想到暴力,但是问题给出的条件太少,所以还是往暴力方向考虑,这种做法自己现在想还是有漏洞,如果不是测评系统给出的数据比较可以的话,1000MS以内应该完不成,以下是代码:#include #incl
kornberg_fresnel
·
2016-02-26 20:00
ACM
poj
ICPC
暴力
POJ 2602 Superlong
sums
(高精度)
Description给出两个长度为n的序列分别表示一个数,输出这两个数的和Input第一行为一整数n表示两序列长度,之后n行每行两个数字(介于0到9之间)分别表示两序列相应位置的数字Output输出一个字符串表示两序列所表示数字相加后的结果SampleInput404426837SampleOutput4750Solution简单题,模拟下加法的进位过程即可,注意此题前置0也要输出Code#in
V5ZSQ
·
2016-02-12 12:00
LeetCode304. Range Sum Query 2D - Immutable
sums
[i][j]=
sums
[i-1][j]+
sums
[i][j-1]-
sums
[i-1][j-1]+matrix[i][j];代码
codeTZ
·
2016-01-26 14:00
LeetCode
动态规划
Backward Digit
Sums
(POJ 3187)
BackwardDigitSumsTimeLimit: 1000MS MemoryLimit: 65536KTotalSubmissions: 5495 Accepted: 3184DescriptionFJandhiscowsenjoyplayingamentalgame.Theywritedownthenumbersfrom1toN(1 2#include 3#include 4#inc
御心飞行
·
2016-01-22 12:00
1125 - Divisible Group
Sums
(DP)
1125-DivisibleGroupSums PDF(English)StatisticsForumTimeLimit:2second(s)MemoryLimit:32MBGivenalistofNnumbersyouwillbeallowedtochooseanyMofthem.SoyoucanchooseinNCMways.Youwillhavetodeterminehowmanyofthe
ZSGG_ACM
·
2016-01-12 16:00
dp
【leetcode】:Range Sum Query - Immutable
这道题非常简单,没啥好说的,直接上代码publicclassNumArray{ int[]
sums
; publicNumArray(int[]nums){
sums
=newint[nums.length
qq_23100787
·
2016-01-07 16:00
LeetCode
Class
poj2140-Herd
Sums
HerdSumsTimeLimit:1000MS MemoryLimit:65536KTotalSubmissions:17068 Accep
qq_33110317
·
2015-12-29 17:00
leetcode -- Range Sum Query 2D - Immutable -- 简单DP题目
构造辅助二维数组sumssums[x][y]表示从0,0到x,y的子矩阵的和利用容斥原理,可知:sumRange(row1,col1,row2,col2)=
sums
[row2][col2]+
sums
[row1
xyqzki
·
2015-12-25 16:00
LeetCode
C#中冒泡排序算法
int[]
sums
={16,62,3,8,5,20};for(inti=0;i<
sums
.Length-1;i++)//总共要比较的次数{for(intj=0;j<
sums
.Length-1-i;j++
汉子的逗比
·
2015-12-23 11:00
POJ 3187 Backward Digit
Sums
(next_permutation()暴力枚举)
BackwardDigitSumsTimeLimit: 1000MSMemoryLimit: 65536KTotalSubmissions: 5412Accepted: 3123DescriptionFJandhiscowsenjoyplayingamentalgame.Theywritedownthenumbersfrom1toN(1 #include #include usingnamespa
zwj1452267376
·
2015-12-15 16:00
poj-2140-Herd
Sums
- 数学规律-连续和为N的方案数
给出n,求 如果 一系列连续的数加起来能恰好得到n,为一种方案,求方案数http://poj.org/problem?id=2140设a,a+1,a+2....a+k=n那么有(k+1)*a+k*(k+1)/2=n;(k+1)*(2*a+k)=2*n如果(2*a+k),k+1是2*n因子;我们只需要找出2*n的所有因子,并判断 (2*a+k)能否被整除即可 #include #include #
viphong
·
2015-11-22 19:00
Leetcode--Combination Sum II
C) and a target number (T), find all unique combinations in C where the candidate numbers
sums
·
2015-11-13 20:00
LeetCode
leetcode------Combination Sum II
C) and a target number (T), find all unique combinations in C where the candidate numbers
sums
·
2015-11-13 14:44
LeetCode
leetcode------Combination Sum
C) and a target number (T), find all unique combinations in C where the candidate numbers
sums
·
2015-11-13 14:43
LeetCode
codechef
Sums
in a Triangle题解
Let's consider a triangle of numbers in which a number appears in the first line, two numbers appear in the second line, three in the third line, etc. Develop a program which will compute the large
·
2015-11-13 14:18
code
Leetcode | Combination Sum I && II
numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers
sums
·
2015-11-13 14:41
LeetCode
LeetCode 040 Combination Sum II
C) and a target number (T), find all unique combinations in C where the candidate numbers
sums
·
2015-11-13 13:39
LeetCode
LeetCode 039 Combination Sum
C) and a target number (T), find all unique combinations in C where the candidate numbers
sums
·
2015-11-13 13:38
LeetCode
USACO Section 2.2: Subset
Sums
dp题,一碰到dp我基本就是跪,搜了网上的答案分两种,一维和二维。 先讲二维,sum[i][j]表示前i个数的subset里差值为j的分法数量。当加入数字i时,有两种选择,某一个set和另外一个set,当加入其中一个总和大的set时,新的差值为j+i,当加入一个总和小的set时,新的差值为abs(j-i)。 1 /* 2 ID: yingzho1 3 LANG: C++ 4 T
·
2015-11-13 13:24
USACO
【BZOJ】1653: [Usaco2006 Feb]Backward Digit
Sums
(暴力)
http://www.lydsy.com/JudgeOnline/problem.php?id=1653 看了题解才会的。。T_T 我们直接枚举每一种情况(这里用next_permutation,全排列) 然后判断是否符合情况(累加判断) #include <cstdio> #include <cstring> #include <cmath>
·
2015-11-13 11:47
USACO
POJ2602-Superlong
sums
转载请注明出处:優YoU http://user.qzone.qq.com/289065406/blog/1305101054 非常恶心的大数相加= = 首先输入就够恶心了。。。哪有人逐位还要间断输入两个数的。。。。 注意: 如果用char[]保存加数和被加数,要用getchar()输入, 如果用int[]保存加数和被加数,要用scanf
·
2015-11-13 10:49
super
leetcode第38题--Combination Sum
C) and a target number (T), find all unique combinations in C where the candidate numbers
sums
·
2015-11-13 08:53
LeetCode
leetcode第39题--Combination Sum II
C) and a target number (T), find all unique combinations in C where the candidate numbers
sums
·
2015-11-13 08:53
LeetCode
LeetCode Online Judge 题目C# 练习 - Combination Sum
numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers
sums
·
2015-11-13 08:44
LeetCode
LeetCode Online Judge 题目C# 练习 - Combination Sum II
numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers
sums
·
2015-11-13 08:44
LeetCode
HDU4193 Non-negative Partial
Sums
(单调队列)
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <queue> 5 using namespace std; 6 const int INF =1<<27; 7 struct node 8 { 9
·
2015-11-13 07:23
part
uva 11997 K Smallest
Sums
首先对每行进行排序,并对与前两行有$A = a_1 \leq a_2 \leq \cdots \leq a_k$和$B = b_1 \leq b_2 \leq \cdots \leq b_k$。首先把所有的$b_i , i\in [1,k]$与$a_1$进行求和,并加入优先队列中。其中最小的必然是$a_1+b_1$,在优先队列队首。然后只需在优先队列中压入$a_1+b_2$,没有压入$a_2+b_
·
2015-11-13 07:52
uva
[LeetCode]Combination Sum II
C) and a target number (T), find all unique combinations in C where the candidate numbers
sums
·
2015-11-13 07:10
LeetCode
[LeetCode]Combination Sum
C) and a target number (T), find all unique combinations in C where the candidate numbers
sums
·
2015-11-13 07:09
LeetCode
leetcode-combination sum and combination sum II
numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers
sums
·
2015-11-13 05:46
LeetCode
学校作业-Usaco DP水题
第一题是USACO 2.2.2 ★Subset
Sums
集合 对于从 1 到 N 的连续整集合合,能划分成两个子集合,且保证每个集合的数字和是相等的.
·
2015-11-13 03:20
USACO
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