hdu5355 Cake(构造)
转载请注明出处: http://www.cnblogs.com/fraud/ ——by fraud
Cake
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1517 Accepted Submission(s): 233
Special Judge
Note that you cannot divide a whole cake into small pieces that is each cake must be complete in the m parts. Each cake must belong to exact one of m parts.
The first contains two integers n and m (1≤n≤105,2≤m≤10), the number of cakes and the number of soda.
It is guaranteed that the total number of soda in the input doesn’t exceed 1000000. The number of test cases in the input doesn’t exceed 1000.
If it is possible, then output m lines denoting the m parts. The first number si of i-th line is the number of cakes in i-th part. Then si numbers follow denoting the size of cakes in i-th part. If there are multiple solutions, print any of them.
其实这题还是不容易过的,比赛的时候一看居然都过了100人了,而且发现直接贪心是不对的,于是乎乱搞了一发。。。结果居然过了。。。
这题首先只有在不能整除以及n<2m-1的时候才是NO,其余情况都是YES
然后在[2m-1,4m-2]的范围内贪心,然后之后,每2m个相互构一组
/**
* code generated by JHelper
* More info: https://github.com/AlexeyDmitriev/JHelper
* @author xyiyy @https://github.com/xyiyy
*/ #include <iostream>
#include <fstream> //#####################
//Author:fraud
//Blog: http://www.cnblogs.com/fraud/
//#####################
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <sstream>
#include <ios>
#include <iomanip>
#include <functional>
#include <algorithm>
#include <vector>
#include <string>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <climits>
#include <cctype> using namespace std;
#define pb(X) push_back(X)
#define rep(X, N) for(int X=0;X<N;X++)
#define rep2(X, L, R) for(int X=L;X<=R;X++)
typedef long long ll; #define gao() out<<"NO"<<endl
vector<int> ans[];
int used[];
ll tot; class hdu5355 {
public:
int ave; void solve(std::istream &in, std::ostream &out) {
int n, m;
in >> n >> m;
tot = (ll) (n + ) * n / ;
if (tot % m != ) {
gao();
return;
}
int tmp = tot / m;
if (n < * m - ) {
gao();
return;
}
int num = n / m;
if (tmp == n) {
out << "YES" << endl;
out << << " " << n << endl;
rep(i, m - ) {
out << << " " << i + << " " << n - i - << endl;
}
return;
}
if (m == ) {
out << "YES" << endl;
out << n;
rep2(i, , n)out << " " << i;
out << endl;
return;
}
rep2(i, , m)ans[i].clear();
int f = n;
int c = ;
/*while(n>=4*m-1){
rep2(i,1,m){
if(c&1) ans[i].pb(f - i + 1);
else ans[i].pb(f - m + i);
}
f -= m;
n -= m;
c++;
rep2(i,1,m){
if(c&1) ans[i].pb(f - i + 1);
else ans[i].pb(f - m + i);
}
f -= m;
n -= m;
c++;
}*/
c = (n + - m * ) % (m * ) + m * - ;
int d = (n - c) / ( * m);
for (int i = , j = c + ; i <= m; i++) {
rep(k, d)ans[i].pb(j++), ans[i].pb(n--);
}
tot = n;
n = c;
ave = (ll) ( + n) * n / / m;
set<int> s;
rep2(i, , n)s.insert(i);
rep2(i, , m) {
rep(j, ave) {
auto it = s.upper_bound(ave - j);
ans[i].pb(*--it);
j += *it;
s.erase(it);
j--;
}
} // clr(used,0);
// rep(i,tot)a[i] = n - i;
// if(dfs(0,0,n,1)){
//if(dfs(m,n,0,0,n)){
out << "YES" << endl;
// rep2(i,1,n)ans[used[i]].pb(i);
rep2(i, , m) {
int sz = ans[i].size();
out << sz;
rep(j, sz) {
out << " " << ans[i][j];
}
out << endl;
}
//}else out<<"NO"<<endl;
/*if(n % m != 0){
gao();
return;
}
if(num&1){
if(m&1){
out<<"YES"<<endl;
run(m);
rep2(i,1,m){
out<<num;
rep(j,ans[i].size()){
out<<" "<<ans[i][j];
}
int last = 3 * m + i;
int f = 4 * m;
rep2(j,4,num){
out<<" "<<last;
f += m;
if(j&1)last = f- m + i;
else last = f - i + 1;
}
out<<endl;
} }else{
gao();
return;
}
}else{
out<<"YES"<<endl;
rep2(i,1,m){
out<<num;
int last = i;
int f = m;
rep(j,num){
out<<" "<<last;
f += m;
if(j&1)last = f-i+1;
else last = f - m + i;
}
}
}*/ } bool dfs(int num, int now, int u, int m) {
if (now == ) {
int i = tot;
while (used[i])i--;
used[i] = m;
if (dfs(num + , i, i - , m))return ;
used[i] = ;
return ;
}
if (now == ave) {
if (num == tot)return ;
else return dfs(num, , tot, m + );
}
for (int i = u; i > ; i--) {
if (!used[i] && now + i <= ave) {
used[i] = m;
if (dfs(num + , now + i, i - , m))return ;
used[i] = ;
}
}
return false;
} bool dfs(int m, int n, int tot, int num, int now) {
if (!m) {
return ;
}
if (tot = ave) {
if (dfs(m - , n, , num, ))return ;
}
if (tot == ) {
int i = n;
while (used[i])i--;
used[i] = m;
if (dfs(m, n, tot + i, num + , i))return ;
used[i] = ;
return ;
}
for (int i = now; i > ; i--) {
if (!used[i] && tot + i <= ave) {
used[i] = m;
if (dfs(m, n, tot + i, num + , i - ))return ;
used[i] = ;
}
}
return false;
} }; int main() {
std::ios::sync_with_stdio(false);
std::cin.tie();
hdu5355 solver;
std::istream &in(std::cin);
std::ostream &out(std::cout);
int n;
in >> n;
for (int i = ; i < n; ++i) {
solver.solve(in, out);
} return ;
}
hdu5355 Cake(构造)的更多相关文章
- hdu 5535 Cake 构造+记忆化搜索
链接:http://acm.hdu.edu.cn/showproblem.php?pid=5355 题意:给定n与m,其中1<= n <= 1e5,2 <= m <= 10;问 ...
- HDU 5355 Cake (构造 + 暴力)
题意:给定 n,m,让你把 1 ~ n 分成 m 部分,而且每部分和是一样大的. 析:首先先判断不能分成的,第一种是 sum (1 ~ n 的和)不能被 m 整除,或者 sum / m < n, ...
- hdu5355 Cake
Problem Description There are m soda and today is their birthday. The 1-st soda has prepared n cakes ...
- HDU 5355 Cake (WA后AC代码,具体解析,构造题)
题目链接:http://acm.hdu.edu.cn/showproblem.php? pid=5355 题面: Cake Time Limit: 2000/1000 MS (Java/Others) ...
- codeforces 1282 E. The Cake Is a Lie (dfs+构造)
链接:https://codeforces.com/contest/1282/problem/E 题意:给的是一张平面图,是一个n边形,每次可以切一刀,切出一个三角形,最终切成n-2个三角形.题目给出 ...
- Java的结构之美【1】——构造对象
当我们遇到多个构造器參数的时候可能会想到用构件器,代码例如以下: /** * 构建器 * @author 阳光小强 * */ public class Lunch { private String c ...
- [Cake] 1. CI中的Cake
在上一篇C#Make自动化构建-简介中,简单的介绍了下Cake的脚本如何编写以及通过Powershell在本地运行Cake脚本.本篇在此基础上,介绍下如何在CI环境中使用Cake. 1. Cake简介 ...
- Linux内核分析— —构造一个简单的Linux系统MenuOS(20135213林涵锦)
Linux内核分析— —构造一个简单的Linux系统MenuOS 实验内容 Linux内核的启动过程,从start_kernel到init进程启动 使用实验楼的虚拟机打开shell cd LinuxK ...
- hdu5355 思维+爆搜
pid=5355">http://acm.hdu.edu.cn/showproblem.php?pid=5355 Problem Description There are m sod ...
随机推荐
- 神奇的match和replace
源自跟奈落大叔的讨论,PHP和JavaScript的比较. 正则: 先说几个正则写法: () 选择匹配一组, (?:) 降低 () 的优先级, .*? 和 .+? ,阻止 . 和 + 的贪婪. 还有一 ...
- GetMemory 函数解析
GetMemory函数 代码1: void GetMemory(char *p){ p = (char*)malloc(100);}int main(int argc, char *argv[]){ ...
- 被误解的 MVC 和被神化的 MVVM
被误解的 MVC 和被神化的 MVVM 被误解的 MVC MVC 的历史 MVC,全称是 Model View Controller,是模型 (model)-视图 (view)-控制器 (contro ...
- Lintcode--009(单词切分)
http://www.lintcode.com/zh-cn/problem/word-break/ 单词切分 给出一个字符串s和一个词典,判断字符串s是否可以被空格切分成一个或多个出现在字典中的单词. ...
- LeetCode_Rotate List
Given a list, rotate the list to the right by k places, where k is non-negative. For example: Given ...
- Borland license information was found,but it is not valid for delphi.
The start Delphi7 come amiss: Borland license information was found,but it is not valid for delphi. ...
- Logstash 介绍
Logstash 介绍: Logstash 是一个开源的数据收集引擎具有实时管道能力, Logstash 可以动态的统一数据从不同的来源和使数据规范化到你选择的目的地. 当Logstash 起初驾驭创 ...
- cf437C The Child and Toy
C. The Child and Toy time limit per test 1 second memory limit per test 256 megabytes input standard ...
- One手动玩转
<preface p2 by Ruiy,我就在开头简单奇葩两句!> 老周被查,涉及到政治问题,我先就不聊了,但Ruiy叹那,都查到七*务了,土党唱哪一出! 能基本玩转OpenNebula都 ...
- URAL 1658
题目大意:求出T个最小的满足各个位的和为S1,平方和为S2的数.按顺序输出.数的位数大于100或者不存在这样一个数时,输出:No solution. KB 64bit IO Format:%I ...