1653: [Usaco2006 Feb]Backward Digit Sums Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 285  Solved: 215[Submit][Status] 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 orde…

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> #include <string> #include <iostream> #include <…

[题目分析] 劳逸结合好了. 杨辉三角+暴搜. [代码] #include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <map> #include <set> #include <queue> #include <string> #include <iostream> #include <a…

每个ai在最后sum中的值是本身值乘上组合数,按这个dfs一下即可 #include<iostream> #include<cstdio> using namespace std; int n,s,ans[15],c[20][20]; bool u[15],f=0; int dfs(int a,int b) { if(b==n) { if(a==s) f=1; return 0; } for(int i=1;i<=n;i++) if(!u[i]) { u[i]=1,ans[b…

1653: [Usaco2006 Feb]Backward Digit Sums Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 207  Solved: 161[Submit][Status][Discuss] Description FJ and his cows enjoy playing a mental game. They write down the numbers from 1 to N (1 <= N <= 10) in a cer…

Backward Digit Sums Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5495   Accepted: 3184 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…

Backward Digit Sums Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5664   Accepted: 3280 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…

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…

POJ 3187  Backward Digit Sums http://poj.org/problem?id=3187 题目大意: 给你一个原始的数字序列: 3   1   2   4  他可以相邻的元素相加得到 4 3 6 然后 7 9 最后得到16,现在给定序列的长度,还有最后的得数,求原始序列(多解则取最小) 思路: 直接枚举即可. 下面是next_permutation版本. #include<cstdio> #include<cstdlib> #include<c…

-->Backward Digit Sums 直接写中文了 Descriptions: FJ 和 他的奶牛们在玩一个心理游戏.他们以某种方式写下1至N的数字(1<=N<=10). 然后把相邻的数相加的到新的一行数.重复这一操作直至只剩一个数字.比如下面是N=4时的一种例子 3 1 2 4 4 3 6 7 9 16 在FJ回来之前,奶牛们开始了一个更难的游戏:他们尝试根据最后结果找到开始的序列.这已超过了FJ的思考极限. 写一个程序来帮助FJ吧 Input N和最后的和 Output 满足…

P1118 [USACO06FEB]Backward Digit Sums G/S 题解:  (1)暴力法.对1-N这N个数做从小到大的全排列,对每个全排列进行三角形的计算,判断是否等于N.  对每个排列进行三角形计算,需要O(N2)次.例如第1行有5个数{a,b,c,d,e},那么第2行计算4次,第3行计算3次-等等,总次数是O(N2)的.  a    b    c    d    e    a+b    b+c   c+d   d+e      a+2b+c b+2c+d c+2d+e   …

题目大意: 给出杨辉三角的顶点值M和底边数的个数 N (1 ≤ N ≤ 10) ,求出底边各个数的值,其中各个数范围都为1 ~ N 当N=4,M=16时可能是这样的   3   1   2   4     4   3   6        7   9            16  Input Multiple test cases. For each case: * Line 1: Two space-separated integers: N and the final sum. Output…

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…

反过来推 题目大意:就是农夫和这只牛又杠上了(怎么老是牛啊,能换点花样吗),给出一行数(从1到N),按杨辉三角的形式叠加到最后,可以得到一个数,现在反过来问你,如果我给你这个数,你找出一开始的序列(可能存在多个序列,输出字典序最小的那个). 这一题首先你要看懂原文的那个1到N是什么意思,就是那一行数只能是1到N,而不是1到10(我一开始犯了这个愚蠢的错误,导致枚举到风扇呼呼的转),如果是这样给你,那么这道题就很简单啦,就直接是用next_permutation枚举所有的序列就可以了,然后找出字典…

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 <…

Description FJ and his cows enjoy playing a mental game. They write down the numbers to N ( <= N <= ) ) might go like Behind FJ's back, the cows have started playing a more difficult game, in which they try to determine the starting sequence from on…

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…

暴力DFS+验证. 验证如果是暴力检验可能复杂度会太高,事实上可以o(1)进行,这个可以o(n*n)dp预处理. #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; ; bool flag[maxn]; bool r; int a[maxn]; int n,sum; ][]; bool check(int s) { ; ; }…

题目:http://poj.org/problem?id=3187 题意: 像这样,输入N : 表示层数,输入over表示最后一层的数字,然后这是一个杨辉三角,根据这个公式,由最后一层的数,推出第一行的数字(由1~N组成).如果有多个解,按字典序升序,输出第一个解. 题解:水题,不多说 AC代码: #include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> usi…

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 number is left. Fo…

第1行j列的一个1加到最后1行满足杨辉三角,可以先推出组合数来 然后next_permutation直接暴. #include<cstdio> #include<iostream> #include <iterator> #include<string> #include<cstring> #include<queue> #include<vector> #include<stack> #include<…

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 number is left. Fo…

题目链接 思路:设一开始的n个数为a1.a2.a3...an,一步一步合并就可以用a1..an表示出最后剩下来的数,不难发现其中a1..an的系数恰好就是第n层杨辉三角中的数.所以我们可以先处理出第n层杨辉三角中的数,然后根据这一层中的数搜索即可. #include<iostream> #include<cstdio> #include<fstream> #include<algorithm> #include<string> #include&…

Backward Digit Sums Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6350   Accepted: 3673 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…

Backward Digit Sums Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5667   Accepted: 3281 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…

1651: [Usaco2006 Feb]Stall Reservations 专用牛棚 Time Limit: 10 Sec  Memory Limit: 64 MBSubmit: 509  Solved: 280[Submit][Status] Description Oh those picky N (1 <= N <= 50,000) cows! They are so picky that each one will only be milked over some precise…

dp( L , R ) = max( dp( L + 1 , R ) + V_L * ( n - R + L ) , dp( L , R - 1 ) + V_R * ( n - R + L ) ) 边界 : dp( i , i ) = V[ i ] * n -------------------------------------------------------------------------------------------- #include<cstdio> #include&l…

线段树.. -------------------------------------------------------------------------------------- #include<cstdio> #include<cstring> #include<algorithm> #include<iostream>   #define rep( i , n ) for( int i = 0 ; i < n ; i++ ) #define…

题目 1652: [Usaco2006 Feb]Treats for the Cows Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 234  Solved: 185[Submit][Status] Description FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for giving vast amounts of milk. FJ se…

题目 1651: [Usaco2006 Feb]Stall Reservations 专用牛棚 Time Limit: 10 Sec  Memory Limit: 64 MBSubmit: 553  Solved: 307[Submit][Status] Description Oh those picky N (1 <= N <= 50,000) cows! They are so picky that each one will only be milked over some preci…