2017 ACM暑期多校联合训练 - Team 5 1008 HDU 6092 Rikka with Subset (找规律)
Problem Description
As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:
Yuta has n positive A1−An and their sum is m. Then for each subset S of A, Yuta calculates the sum of S.
Now, Yuta has got 2n numbers between [0,m]. For each i∈[0,m], he counts the number of is he got as Bi.
Yuta shows Rikka the array Bi and he wants Rikka to restore A1−An.
It is too difficult for Rikka. Can you help her?
Input
The first line contains a number t(1≤t≤70), the number of the testcases.
For each testcase, the first line contains two numbers n,m(1≤n≤50,1≤m≤104).
The second line contains m+1 numbers B0−Bm(0≤Bi≤2n).
Output
For each testcase, print a single line with n numbers A1−An.
It is guaranteed that there exists at least one solution. And if there are different solutions, print the lexicographic minimum one.
Sample Input
2
2 3
1 1 1 1
3 3
1 3 3 1
Sample Output
1 2
1 1 1
Hint
In the first sample, \(A\) is \([1,2]\). \(A\) has four subsets \([],[1],[2],[1,2]\) and the sums of each subset are \(0,1,2,3\). So \(B=[1,1,1,1]\)
题意:
给定数组b,保存的数组a中的所有子集和的个数,让找个这个数组a,并且按照字典序输出来。
分析:
b数组里面除了第一个元素b[0]肯定为1之外(表示空集的个数有且仅有一个),其余的第一次出现的b[i]==j(j!=0),那么就表示i这个数字肯定在a数组中存在,而且为第一个并且为最小的元素,同时将这个数的个数减减(相当于减去单独自己本身一个自己的情况),记录下标i。
这样循环的往后加b数组的元素下标偏移i个单位,如果此时两个数的个数均不为0,也就意味这后面的那个数,可以由前面这个数构成,然后让当前下标的数减去b[i],得到的那个数减去b[i]的个数,(相当于减去这个数可以由b[i]组合而成的个数)剩下的肯定就是这个数字本身的个数
就这样循环着往下找
代码:
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxm = 1100002;
int b[maxm];
int m;
int a[maxm];
int main()
{
int t;
scanf( "%d", &t );
while( t-- )
{
int n;
scanf( "%d%d", &n, &m );
for( int i = 0; i <= m; i++ )
{
scanf( "%d", b+i );
}
b[0] = 0;///空集的情况,不用任何的元素构成,将其赋值为0,不影响后面的计算
for( int i = 0; i < n; i++ )///循环得到a数组中的第i个元素
{
int j;
for( j = 0; j <= m; j++ )///找到当前的第一个不为0的
{
if( b[j] ) break;
}
b[j]--;///相当于有一个本身构成的元素
a[i] = j;///那么就肯定有一个这个元素,而且为当前的最小值
for( int k = j; k <= m; k++ )///找后面的所有可以由j组合而成的是数
{
if( k+j <= m && b[k] && b[k+j] )
{
b[k+j] -= b[k];
}
}
}
for( int i = 0; i < n; i++ )
{
i == 0 ? printf( "%d", a[i] ) : printf( " %d", a[i] );
}
printf( "\n" );
}
return 0;
}
2017 ACM暑期多校联合训练 - Team 5 1008 HDU 6092 Rikka with Subset (找规律)的更多相关文章
- 2017 ACM暑期多校联合训练 - Team 9 1008 HDU 6168 Numbers (模拟)
题目链接 Problem Description zk has n numbers a1,a2,...,an. For each (i,j) satisfying 1≤i<j≤n, zk gen ...
- 2017 ACM暑期多校联合训练 - Team 3 1008 HDU 6063 RXD and math (莫比乌斯函数)
题目链接 Problem Description RXD is a good mathematician. One day he wants to calculate: ∑i=1nkμ2(i)×⌊nk ...
- 2017 ACM暑期多校联合训练 - Team 4 1007 HDU 6073 Matching In Multiplication (模拟)
题目链接 Problem Description In the mathematical discipline of graph theory, a bipartite graph is a grap ...
- 2017 ACM暑期多校联合训练 - Team 4 1012 HDU 6078 Wavel Sequence (模拟)
题目链接 Problem Description Have you ever seen the wave? It's a wonderful view of nature. Little Q is a ...
- 2017ACM暑期多校联合训练 - Team 7 1010 HDU 6129 Just do it (找规律)
题目链接 Problem Description There is a nonnegative integer sequence a1...n of length n. HazelFan wants ...
- 2017ACM暑期多校联合训练 - Team 2 1006 HDU 6050 Funny Function (找规律 矩阵快速幂)
题目链接 Problem Description Function Fx,ysatisfies: For given integers N and M,calculate Fm,1 modulo 1e ...
- 2017ACM暑期多校联合训练 - Team 8 1008 HDU 6140 Hybrid Crystals (模拟)
题目链接 Problem Description Kyber crystals, also called the living crystal or simply the kyber, and kno ...
- 2017ACM暑期多校联合训练 - Team 7 1008 HDU 6127 Hard challenge (极角排序)
题目链接 Problem Description There are n points on the plane, and the ith points has a value vali, and i ...
- 2017ACM暑期多校联合训练 - Team 6 1008 HDU 6103 Kirinriki (模拟 尺取法)
题目链接 Problem Description We define the distance of two strings A and B with same length n is disA,B= ...
随机推荐
- QSerialPort-Qt串口通讯
版权声明:若无来源注明,Techie亮博客文章均为原创. 转载请以链接形式标明本文标题和地址: 本文标题:QSerialPort-Qt串口通讯 本文地址:http://techieliang. ...
- VisualStudio2013 代码查看优化 对齐线
http://jingyan.baidu.com/article/363872eccef5276e4ba16f91.html
- 一些Redis面试题
1. 使用Redis有哪些好处? (1) 速度快,因为数据存在内存中,类似于HashMap,HashMap的优势就是查找和操作的时间复杂度都是O(1) (2) 支持丰富数据类型,支持string,li ...
- PHP中大括号用法
Php中"{}"大括号的用法总结 在PHP中,大括号“{}”可以起到如下作用: 1.将多个独立语句合并为一个复合语句,例如 if ... else ...中经常如此使用 2.在变量 ...
- vue.cli实现tab切换效果
<template> <div class="cp-select"> <div class="lef ...
- 【uoj#51】[UR #4]元旦三侠的游戏 博弈论+dp
题目描述 给出 $n$ 和 $m$ ,$m$ 次询问.每次询问给出 $a$ 和 $b$ ,两人轮流选择:将 $a$ 加一或者将 $b$ 加一,但必须保证 $a^b\le n$ ,无法操作者输,问先手是 ...
- 使用Runtime.getRuntime().exec()方法的几个陷阱
Process 子类的一个实例,该实例可用来控制进程并获得相关信息.Process 类提供了执行从进程输入.执行输出到进程.等待进程完成.检查进程的退出状态以及销毁(杀掉)进程的方法. 创建进程的方法 ...
- spring的事务传播特性
PROPAGATION_REQUIRED(常用) Support a current transaction; create a new one if none exists. 支持一个当前事务;如 ...
- 【题解】JSOI2011分特产
没sa可suo的,sui题一道…… #include <bits/stdc++.h> using namespace std; #define maxn 3000 #define mod ...
- 跟我学Spring Cloud(Finchley版)-20-Spring Cloud Config-Git仓库配置详解 原
在跟我学Spring Cloud(Finchley版)-19-配置中心-Spring Cloud Config 一节中,已实现使用Git仓库作为Config Server的后端存储,本节详细探讨如何配 ...