最小的差别 题目大意:输入一串数字,问你数字的组合方式,你可以随意用这些数字组合(无重复)来组合成一些整数(第一位不能为0),然后问你组合出来的整数的差最小是多少? 这一题可以用枚举的方法去做,这里我就用枚举和贪心(这个非常爽)的方法去. 首先这一题的限制是1000ms,我们知道全枚举的时间肯定是超时的,所以我们必须裁枝,我们这样看,我们得到的最小值,必须是两个数差别最小的时候才会出现,所以这里我们可以只用枚举lenth/2时候的数串的情况,这样就降低了很多的复杂度. 但是这样还是不够,但是我们…

 Smallest Difference(最小差) Time Limit: 1000MS    Memory Limit: 65536K Description - 题目描述 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 d…

Smallest Difference Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6740   Accepted: 1837 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…

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 remaining digits can be written down in some order to form a second…

-->Smallest Difference 直接写中文了 Descriptions: 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  输入第一行的数表示随后测试用例的数量.对于每组测试用例,有一行至少两个…

Smallest Difference Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6493   Accepted: 1771 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…

Smallest Difference Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 19528   Accepted: 5329 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…

Binary search. class Solution { int _findClosest(vector<int> &A, int v) { , e = A.size() - ; int ret = INT_MAX; while(s <= e) { ; int vmid = A[mid]; int dist = abs(vmid - v); ret = min(ret, dist); ; if(vmid < v) { s = mid + ; } else if(vmi…

POJ 2718 问题描述: 给一串数,求划分后一个子集以某种排列构成一个数,余下数以某种排列构成另一个数,求这两个数最小的差,注意0开头的处理. 超时问题:一开始是得到一个数列的组合之后再从中间进行切割得到两数,会超时.后来采用的方法是将前面的数在DFS中得到固定,在函数work中对后面(n-n/2)个数进行排列组合. 针对整个数列的dfs排列组合剪枝,若当前搜索的第二个数后面全部补零与第一个数所产生的差值比当前所搜索到的结果还要大,那么就直接返回.这个剪枝就是超时与几十MS的差距int nu…

Given two array of integers(the first array is array A, the second array is arrayB), 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 that the difference between A[i] and B[j] (|A[i] - B[j]…