POJ2718 Smallest Difference Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6509   Accepted: 1773 Description Given a number of distinct decimal digits, you can form one integer by choosing a non-empty subset of these digits and writing…

题目链接: https://vjudge.net/problem/POJ-3187 题目大意: 输入n,sum,求1~n的数,如何排列之后,相邻两列相加,直到得出最后的结果等于sum,输出1~n的排列(杨辉三角)  3 1 2 4 //1~n 全排列中的一个排列  4 3 6  7 9 sum = 16 思路: 直接枚举1-n的全排列,判断之和是不是sum #include<iostream> #include<vector> #include<queue> #incl…

题目地址 简要题意: 输入两个数n和m,分别表示给你1--n这些整数,将他们按一定顺序摆成一行,按照杨辉三角的计算方式进行求和,求使他们求到最后时结果等于m的排列中字典序最小的一种. 思路分析: 不难推得第一行为n个数a1\a2\--\an时求得的和为i=0∑n-1 ai*(n-1Ci) 根据此公式,考虑到数据量比较小,只需要将原本按递增顺序依次排列好的1--n按next_permutation给出的递增全排列顺序逐个代入,如果结果与m相等就停止循环即可. 参考代码: #include<stdi…

http://poj.org/problem?id=3187 给定一个个数n和sum,让你求原始序列,如果有多个输出字典序最小的. 暴力枚举题,枚举生成的每一个全排列,符合即退出. dfs版: #include <iostream> #include <cstdio> #include <cmath> #include <vector> #include <cstring> #include <string> #include <…

给出杨辉三角的顶点值,求底边各个数的值.直接DFS就好了 #include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<queue> #include<set> #include<map> #include<vector> #include<cmath&g…

一.题目 Description FJ and his cows enjoy playing a mental game. They write down the numbers from 1 to N (1 <= N <= 10) in a certain order and then sum adjacent numbers to produce a new list with one fewer number. They repeat this until only a single n…

Backward Digit Sums Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4487   Accepted: 2575 Description FJ and his cows enjoy playing a mental game. They write down the numbers from 1 to N (1 <= N <= 10) in a certain order and then sum ad…

一.题意:给定n,求1~n的一个排列,这个排列需要满足以下两个要求:1.杨辉三角最后的和为sum  2.字典序最小 二.思路:暴力枚举每一个排列,然后计算和并与sum进行比较.这里我比较费解的是为什么没超时... 三.代码: #include"iostream" #include"stdio.h" #include"algorithm" using namespace std; int n,sum; int GetArraySum(int *a)…

Backward Digit Sums Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7634   Accepted: 4398 Description FJ and his cows enjoy playing a mental game. They write down the numbers from 1 to N (1 <= N <= 10) in a certain order and then sum ad…

挑战-搜索 题意: 给一个n和sum,代表n层的杨辉三角,然后给一个和,问最低层的数字情况. 思路: ①:预处理一个底层对于和的系数数组, sum = 0Cn-1*num[1] + 1Cn-1*num[2] +-+ n-1Cn-1*num[n]; ②:因为底层就是1-n直接暴搜-即可- 贴一发挫code--- #include<cstdio> #include<map> #include<queue> #include<math.h> #include<…