Smallest Difference

最新推荐文章于 2020-05-06 10:37:47 发布
最新推荐文章于 2020-05-06 10:37:47 发布
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这是一个关于寻找给定整数集合中两个数之差最小的问题。输入包含一组整数，任务是找到其中差值最小的两个数并返回这个差值。

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Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 5364
Accepted: 1456

Description

Given a number of distinct decimal digits, you can form
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（全排列）Smallest Difference （poj2718）
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题目： Description - 题目描述 给定若干位十进制数，你可以通过选择一个非空子集并以某种顺序构建一个数。剩余元素可以用相同规则构建第二个数。除非构造的数恰好为0，否则不能以0打头。 举例来说，给定数字0，1，2，4，6与7，你可以写出10和2467。当然写法多样：210和764，204和176，等等。最后一对数差的绝对值为28，实际上没有其他对拥有更小的差。 Input - ...
Smallest Difference（枚举）
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Smallest Difference Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3973   Accepted: 1110 Description Given a number of distinct decimal digits, you can form one i...
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D - Smallest Difference Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Description Given a number of distinct decimal digits, you can form
【题解】POJ 2718 Smallest Difference(搜索)
bu_shushu的博客
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POJ 2718 Smallest Difference 原题 https://vjudge.net/problem/POJ-2718 Given a number of distinct decimal digits, you can form one integer by choosing a non-empty subset of these digits and writing th...
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Load balancing (负载均衡) refers to efficiently distributing incoming network traffic across a group of backend servers. A load balancing algorithm distributes loads in a specific way. If we can estimate the maximum incoming traffic load, here is an algorithm that works according to the following rule: The incoming traffic load of size S will first be partitioned into two parts, and each part may be again partitioned into two parts, and so on. Only one partition is made at a time. At any time, the size of the smallest load must be strictly greater than half of the size of the largest load. All the sizes are positive integers. This partition process goes on until it is impossible to make any further partition. For example, if S=7, then we can break it into 3+4 first, then continue as 4=2+2. The process stops at requiring three servers, holding loads 3, 2, and 2. Your job is to decide the maximum number of backend servers required by this algorithm. Since such kind of partitions may not be unique, find the best solution -- that is, the difference between the largest and the smallest sizes is minimized. Input Specification: Each input file contains one test case, which gives a positive integer S (2≤N≤200), the size of the incoming traffic load. Output Specification: For each case, print two numbers in a line, namely, M, the maximum number of backend servers required, and D, the minimum of the difference between the largest and the smallest sizes in a partition with M servers. The numbers in a line must be separated by one space, and there must be no extra space at the beginning or the end of the line. Sample Input: 22 Sample Output: 4 1 Hint: There are more than one way to partition the load. For example: 22 = 8 + 14 = 8 + 7 + 7 = 4 + 4 + 7 + 7 or 22 = 10 + 12 = 10 + 6 + 6 = 4 + 6 + 6 + 6 or 22 = 10 + 12 = 10 + 6 + 6 = 5 + 5 + 6 + 6 All requires 4 servers. The last partition has the smallest difference 6−5=1, hence 1 is printed out.请给这一到题设置测试用例，包含最大最小情况，和一些特殊的情况，以及对于回溯算法最好和最坏的情况
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嗯，用户需要为负载均衡算法设计测试用例，覆盖边界情况、特殊分区以及回溯算法的性能评估。首先，我得理清楚负载均衡算法的常见输入范围和关键参数。根据用户提到的，输入值的最小是2，最大是200，所以边界值测试...
The Smallest Difference
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Given two array of integers(the first array is array A, the second array is array B), now we are going to find a element in array A which is A[i], and another element in array B which is B[j], so
D - Smallest Difference
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Smallest Difference Given a number of distinct decimal digits, you can form one integer by choosing a non-empty subset of these digits and writing them in some order. The remaining digits can b
POJ2718 UVALive3341 Smallest Difference【全排列+暴力】
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Smallest DifferenceTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 15423 Accepted: 4173DescriptionGiven a number of distinct decimal digits, you can form one integer by choosing a non-empty ...
Smallest Difference-模拟
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#include using namespace std; int main() {     int t;     scanf("%d\n",&t);     while (t--) {         char a[20];         gets(a);         int i,j,flag=15,len=0,k=strlen(a),b[20]={0},m
Smallest Difference(dfs+STL)
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Problem:Smallest Difference Description: Given a number of distinct decimal digits, you can form one integer by choosing a non-empty subset of these digits and writing them in some order. The remainin...
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Given a number of distinct decimal digits, you can form one integer by choosing a non-empty subset of these digits and writing them in some order. The remaining digits can be written down in some orde
Smallest Difference (POJ 2718, 穷竭搜索）
c138Rick的博客
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这个题有两种做法：1.使用next_permutation穷举 2.贪心法        这道题让我发现了一个问题，执行pow(x,y)这个函数就这么费时间么···        因为这个题在书中属于穷竭搜索的练习题，这里就不说贪心法怎么做了。解题思路就是排列，然后分成两组数，计算差。其实非常无脑，但是我提交了很多遍都TLE。。。直到我看了一下别人写的代码，发现在将一组数字组合成一个多位数时别
POJ-2718-Smallest Difference(next_permutation)
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Smallest Difference Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 13820 Accepted: 3739 Description Given a number of distinct decimal digits, you can form one integer by c
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