A

实际上我们for一遍就好。

坑点就是会爆int

#include<bits/stdc++.h>

using namespace std;

const int maxn = 1e5+7;

long long ans1,ans2,a[maxn],b[maxn],c[maxn];

int n;

int main(){

    while(cin>>n){

        ans1 = -1;

        for(int i=1;i<=n;i++){

            cin>>a[i]>>b[i]>>c[i];

            if(a[i]*b[i]*c[i]>ans1){

                ans1 = a[i]*b[i]*c[i];

                ans2 = i;

            }

        }

        cout<<ans2<<endl;

    }

}

B . 两数组找相同的元素

我们对于B的每一个元素，在A中进行二分判断是否存在即可。

也可以使用C++内带的.find函数

#include <cstdio>

#include <unordered_set>

#include <vector>

using namespace std;

int main(int argc, const char *argv[])

{

    unordered_set<int> first_array;

    int m, n;

    scanf("%d %d", &m, &n);

    int tmp;

    for (int i = 0; i < m; i++) {

        scanf("%d", &tmp);

        first_array.emplace(tmp);

    }

    vector<int> ans;

    for (int i = 0; i < n; i++) {

        scanf("%d", &tmp);

        if (first_array.find(tmp) != first_array.end()) {

            ans.push_back(tmp);

        }

    }

    for (int i = 0; i < ans.size(); ++i) {

        printf("%d%c", ans[i], i + 1 == ans.size() ? '\n' : ' ');

    }

    return 0;

}

B . DAU统计

实际上我们对于每一个元素，如果这个元素之前并没有出现过，那么对答案的贡献就加一。

于是我们可以用hash每次O（1）的判断这个元素之前是否出现过即可。

也可以使用set去维护也可以。

// Copyright 2017 He Tianyi <hetianyi@bytedance.com>

// Compiler opts: g++ -g -std=c++11 -o dau -O2 -Wall -Werror dau.cc

#include <cstdint>

#include <iostream>

#include <vector>

#include <set>

#include <algorithm>

#include <memory>

static const size_t kMergeThreshold = 170000;

class UserMap {

 public:

  UserMap() {

    base_ = std::unique_ptr<std::vector<uint64_t>>(

        new std::vector<uint64_t>());

  }

  void Merge() {

    std::vector<uint64_t> *result = new std::vector<uint64_t>();

    result->reserve(base_->size() + delta_.size());

    auto base_it = base_->begin();

    auto delta_it = delta_.begin();

    while (base_it != base_->end() || delta_it != delta_.end()) {

      if (delta_it == delta_.end()) {

        result->emplace_back(*base_it);

        base_it++;

      } else if (base_it == base_->end()) {

        result->emplace_back(*delta_it);

        delta_it++;

      } else {

        if (*base_it < *delta_it) {

          result->emplace_back(*base_it);

          base_it++;

        } else {

          result->emplace_back(*delta_it);

          delta_it++;

        }

      }

    }

    base_.reset(result);

    delta_.clear();

  }

  bool PutIfAbsent(uint64_t uid) {

    if (delta_.find(uid) != delta_.end()) {

      return false;

    }

    if (std::binary_search(base_->begin(), base_->end(), uid)) {

      return false;

    }

    delta_.insert(uid);

    if (delta_.size() > kMergeThreshold) {

      Merge();

    }

    return true;

  }

 private:

  std::unique_ptr<std::vector<uint64_t>> base_;

  std::set<uint64_t> delta_;

};

int main(int argc, char** argv) {

  uint64_t uid;

  UserMap m;

  uint32_t result = 0;

  do {

    std::cin >> uid;

    if (uid > 0 && m.PutIfAbsent(uid)) {

      result++;

    }

  } while (uid != 0);

  std::cout << result << std::endl;

  return 0;

}

D . 形式化计算

题面分为两个部分，计算部分和输出部分。

计算部分比较简单，我们对于每个符号进行判断即可。

计算部分结束之后，我们要处理输出。

输出我们需要对每个数字/符号单独输出。可以观察到，每个字符实际上都是一个矩阵，所以我们可以预处理出每个字符矩阵的样子，然后输出即可。

#include <iostream>

#include <cstdio>

#include <cstring>

using namespace std;

int a,b;

char s[10];

char drawBoard[5][200];

void print1(int &startCol) {

	drawBoard[0][startCol] = '*';

	drawBoard[1][startCol] = '*';

	drawBoard[2][startCol] = '*';

	drawBoard[3][startCol] = '*';

	drawBoard[4][startCol] = '*';

	startCol = startCol + 3;

}

void print2(int &startCol) {

	drawBoard[0][startCol] = '*';

	drawBoard[0][startCol+1] = '*';

	drawBoard[0][startCol+2] = '*';

	drawBoard[1][startCol+2] = '*';

	drawBoard[2][startCol+2] = '*';

	drawBoard[2][startCol+1] = '*';

	drawBoard[2][startCol] = '*';

	drawBoard[3][startCol] = '*';

	drawBoard[4][startCol] = '*';

	drawBoard[4][startCol+1] = '*';

	drawBoard[4][startCol+2] = '*';

	startCol = startCol + 5;

}

void print3(int &startCol) {

	drawBoard[0][startCol] = '*';

	drawBoard[0][startCol+1] = '*';

	drawBoard[0][startCol+2] = '*';

	drawBoard[1][startCol+2] = '*';

	drawBoard[2][startCol] = '*';

	drawBoard[2][startCol+1] = '*';

	drawBoard[2][startCol+2] = '*';

	drawBoard[3][startCol+2] = '*';

	drawBoard[4][startCol] = '*';

	drawBoard[4][startCol+1] = '*';

	drawBoard[4][startCol+2] = '*';

	startCol = startCol + 5;

}

void print4(int &startCol) {

	drawBoard[0][startCol] = '*';

	drawBoard[0][startCol+2] = '*';

	drawBoard[1][startCol] = '*';

	drawBoard[1][startCol+2] = '*';

	drawBoard[2][startCol] = '*';

	drawBoard[2][startCol+1] = '*';

	drawBoard[2][startCol+2] = '*';

	drawBoard[3][startCol+2] = '*';

	drawBoard[4][startCol+2] = '*';

	startCol = startCol + 5;

}

void print5(int &startCol) {

	drawBoard[0][startCol] = '*';

	drawBoard[0][startCol+1] = '*';

	drawBoard[0][startCol+2] = '*';

	drawBoard[1][startCol] = '*';

	drawBoard[2][startCol] = '*';

	drawBoard[2][startCol+1] = '*';

	drawBoard[2][startCol+2] = '*';

	drawBoard[3][startCol+2] = '*';

	drawBoard[4][startCol+2] = '*';

	drawBoard[4][startCol+1] = '*';

	drawBoard[4][startCol] = '*';

	startCol = startCol + 5;

}

void print6(int &startCol) {

	drawBoard[0][startCol] = '*';

	drawBoard[0][startCol+1] = '*';

	drawBoard[0][startCol+2] = '*';

	drawBoard[1][startCol] = '*';

	drawBoard[2][startCol] = '*';

	drawBoard[2][startCol+1] = '*';

	drawBoard[2][startCol+2] = '*';

	drawBoard[3][startCol] = '*';

	drawBoard[3][startCol+2] = '*';

	drawBoard[4][startCol] = '*';

	drawBoard[4][startCol+1] = '*';

	drawBoard[4][startCol+2] = '*';

	startCol = startCol + 5;

}

void print7(int &startCol) {

	drawBoard[0][startCol] = '*';

	drawBoard[0][startCol+1] = '*';

	drawBoard[0][startCol+2] = '*';

	drawBoard[1][startCol+2] = '*';

	drawBoard[2][startCol+2] = '*';

	drawBoard[3][startCol+2] = '*';

	drawBoard[4][startCol+2] = '*';

	startCol = startCol + 5;

}

void print8(int &startCol) {

	drawBoard[0][startCol] = '*';

	drawBoard[0][startCol+1] = '*';

	drawBoard[0][startCol+2] = '*';

	drawBoard[1][startCol] = '*';

	drawBoard[1][startCol+2] = '*';

	drawBoard[2][startCol] = '*';

	drawBoard[2][startCol+1] = '*';

	drawBoard[2][startCol+2] = '*';

	drawBoard[3][startCol] = '*';

	drawBoard[3][startCol+2] = '*';

	drawBoard[4][startCol] = '*';

	drawBoard[4][startCol+1] = '*';

	drawBoard[4][startCol+2] = '*';

	startCol = startCol + 5;

}

void print9(int &startCol) {

	drawBoard[0][startCol] = '*';

	drawBoard[0][startCol+1] = '*';

	drawBoard[0][startCol+2] = '*';

	drawBoard[1][startCol] = '*';

	drawBoard[1][startCol+2] = '*';

	drawBoard[2][startCol] = '*';

	drawBoard[2][startCol+1] = '*';

	drawBoard[2][startCol+2] = '*';

	drawBoard[3][startCol+2] = '*';

	drawBoard[4][startCol] = '*';

	drawBoard[4][startCol+1] = '*';

	drawBoard[4][startCol+2] = '*';

	startCol = startCol + 5;

}

void print0(int &startCol) {

	drawBoard[0][startCol] = '*';

	drawBoard[0][startCol+1] = '*';

	drawBoard[0][startCol+2] = '*';

	drawBoard[1][startCol] = '*';

	drawBoard[1][startCol+2] = '*';

	drawBoard[2][startCol] = '*';

	drawBoard[2][startCol+2] = '*';

	drawBoard[3][startCol] = '*';

	drawBoard[3][startCol+2] = '*';

	drawBoard[4][startCol] = '*';

	drawBoard[4][startCol+1] = '*';

	drawBoard[4][startCol+2] = '*';

	startCol = startCol + 5;

}

void printjia(int &startCol) {

	drawBoard[1][startCol+1] = '*';

	drawBoard[2][startCol] = '*';

	drawBoard[2][startCol+1] = '*';

	drawBoard[2][startCol+2] = '*';

	drawBoard[3][startCol+1] = '*';

	startCol = startCol + 5;

}

void printjian(int &startCol) {

	drawBoard[2][startCol] = '*';

	drawBoard[2][startCol+1] = '*';

	drawBoard[2][startCol+2] = '*';

	startCol = startCol + 5;

}

void printcheng(int &startCol) {

	drawBoard[1][startCol] = '*';

	drawBoard[1][startCol+2] = '*';

	drawBoard[2][startCol+1] = '*';

	drawBoard[3][startCol] = '*';

	drawBoard[3][startCol+2] = '*';

	startCol = startCol + 5;

}

void printchu(int &startCol) {

	drawBoard[1][startCol+2] = '*';

	drawBoard[2][startCol+1] = '*';

	drawBoard[3][startCol] = '*';

	startCol = startCol + 5;

}

void printequal(int &startCol) {

	drawBoard[1][startCol] = '*';

	drawBoard[1][startCol+1] = '*';

	drawBoard[1][startCol+2] = '*';

	drawBoard[1][startCol+3] = '*';

	drawBoard[3][startCol] = '*';

	drawBoard[3][startCol+1] = '*';

	drawBoard[3][startCol+2] = '*';

	drawBoard[3][startCol+3] = '*';

	startCol = startCol + 6;

}

void printdian(int &startCol) {

	drawBoard[3][startCol] = '*';

	drawBoard[3][startCol+1] = '*';

	drawBoard[4][startCol] = '*';

	drawBoard[4][startCol+1] = '*';

	startCol = startCol + 4;

}

void drawNum(double x,int &startCol) {

	char s[100];

	memset(s,0,sizeof(s));

	if (x-int(x) == 0) sprintf(s,"%d",int(x));

	else if (x*10-int(x*10)==0) sprintf(s,"%.1lf",x);

	else sprintf(s,"%.2lf",x);

	for (int i=0;i<strlen(s);++i) {

		switch (s[i]) {

			case '0':print0(startCol);break;

			case '1':print1(startCol);break;

			case '2':print2(startCol);break;

			case '3':print3(startCol);break;

			case '4':print4(startCol);break;

			case '5':print5(startCol);break;

			case '6':print6(startCol);break;

			case '7':print7(startCol);break;

			case '8':print8(startCol);break;

			case '9':print9(startCol);break;

			case '.':printdian(startCol);break;

			case '-':printjian(startCol);break;

		}

	}

}

int main() {

	scanf("%d %s %d",&a,s,&b);

	for (int i=0;i<5;++i)

		for (int j=0;j<200;++j)

			drawBoard[i][j] = ' ';

	double c=0;

	int startCol=0;

	drawNum(a,startCol);

	switch (s[0]) {

		case '+':c = a+b; printjia(startCol);break;

		case '-':c = a-b; printjian(startCol);break;

		case '*':c = a*b; printcheng(startCol);break;

		case '/':c = 1.0*a/b; printchu(startCol);break;

	}

	drawNum(b,startCol);

	printequal(startCol);

	drawNum(c,startCol);

	for (int i=0;i<5;++i,cout<<endl)

		for (int j=0;j<startCol;++j)

			cout<<drawBoard[i][j];

	return 0;

}

E . 任务执行策略

动态规划题。

首先把三角形翻转一下，倒着来考虑每个元素。

dp[i][j][k]表示考虑到第(i,j)个，当前选取了k个元素的最大值。

前缀和维护最大值，就可以把转移优化到O(1)。

#include <algorithm>

#include <cassert>

#include <cstring>

#include <cstdio>

const int N = 60;

const int M = 500 + 10;

int dp[N][N][M], sum[N][N], a[N][N], n, m;

int main() {

    assert(scanf("%d%d", &n, &m) == 2);

    assert(1 <= n && n <= 50);

    assert(1 <= m && m <= 500);

    for (int i = 1; i <= n; ++ i) {

        for (int j = 1; j <= i; ++ j) {

            assert(scanf("%d", &a[i][j]) == 1);

            assert(0 <= a[i][j] && a[i][j] <= 1000);

        }

    }

    for (int i = 1; i <= n; ++ i) {

        for (int j = 1; j <= i; ++ j) {

            sum[i][j] = sum[i][j - 1] + a[n - j + 1][i - j + 1];

        }

    }

    memset(dp, 200, sizeof(dp));

    for (int i = 0; i <= n; ++ i) {

        dp[i][0][0] = 0;

    }

    for (int i = 1; i <= n; ++ i) {

        for (int j = i; j >= 0; -- j) {

            for (int k = j; k <= m; ++ k) {

                dp[i][j][k] = std::max(dp[i][j + 1][k],

                                       dp[i - 1][std::max(0, j - 1)][k - j] + sum[i][j]);

            }

        }

    }

    printf("%d\n", dp[n][0][m]);

    return 0;

}

喵哈哈村的魔法考试 Round 16 (Div.2) 比赛题解的更多相关文章

喵哈哈村的魔法考试 Round &num;7 (Div&period;2) 题解
喵哈哈村的魔法考试 Round #7 (Div.2) 注意!后四道题来自于周日的hihocoder offer收割赛第九场. 我建了个群:欢迎加入qscoj交流群,群号码:540667432 大概作为 ...
喵哈哈村的魔法考试 Round &num;2 (Div&period;2) 题解
喵哈哈村的魔法考试 Round #2 (Div.2) 题解 A.喵哈哈村的战争题解: 这道题就是for一遍,统计每个村子的战斗力的和,然后统计哪个村子的战斗力和大一点就好了. 唯一的坑点,就是这道题 ...
喵哈哈村的魔法考试 Round &num;1 (Div&period;2) 题解
喵哈哈村的魔法考试 Round #1 (Div.2) 题解特别感谢出题人,qscqesze. 也特别感谢测题人Xiper和CS_LYJ1997. 没有他们的付出,就不会有这场比赛. A 喵哈哈村的魔 ...
喵哈哈村的魔法考试 Round &num;1 (Div&period;2) 题解&amp&semi;源码(A&period;水+暴力，B&period;dp+栈)
A.喵哈哈村的魔法石发布时间: 2017年2月21日 20:05 最后更新: 2017年2月21日 20:06 时间限制: 1000ms 内存限制: 128M 描述传说喵哈哈村有三种神 ...
喵哈哈村的魔法考试 Round &num;19 (Div&period;2) 题解
题解: 喵哈哈村的魔力源泉(1) 题解:签到题. 代码: #include<bits/stdc++.h> using namespace std; int main(){ long lon ...
喵哈哈村的魔法考试 Round &num;1 (Div&period;2)
比赛地址:http://qscoj.cn/contest/2/ 都是中文题,这里不在详述题意 A.喵哈哈村的魔法石分析:暴力求解 #include<iostream> #include& ...
喵哈哈村的魔法考试 Round &num;1 (Div&period;2) ABCD
官方题解: http://www.cnblogs.com/qscqesze/p/6418555.html#3623453 喵哈哈村的魔法石描述传说喵哈哈村有三种神奇的魔法石:第一种魔法石叫做人铁石 ...
喵哈哈村的魔法考试 Round &num;4 (Div&period;2) 题解
有任何疑问,可以加我QQ:475517977进行讨论. A 喵哈哈村的嘟嘟熊魔法(1) 题解这道题我们只要倒着来做就可以了,因为交换杯子是可逆的,我们倒着去模拟一遍就好了. 有个函数叫做swap(a ...
喵哈哈村的魔法考试 Round &num;20 (Div&period;2) 题解
题解: A 喵哈哈村的跳棋比赛题解:其实我们要理解题意就好了,画画图看看这个题意.x<y,那么就交换:x>y,那么x=x%y. 如果我们经过很多次,或者y<=0了,那么就会无限循环 ...

随机推荐

考前预习（Ubuntu配备）
这几天考前预习,趁现在不想预习,写点之前就想写的东西吧. 贴一下个人认为有用的,在Ubuntu装机后的一些小事.不过挺杂的,主要是拿来给以后的自己看,以及让现在无聊的我有点事做. 首先,Ubuntu官 ...
不等高cell搭建(二)
一.commentView模块搭建 commentView样式分为两种 1.xib搭建界面 1.1 因为评论的样式大体上一样,我们可以用同一个xib来处理 1.2 最热评论用 ...
设置JVM参数，查看堆大小
1.在eclipse设置JVM参数打开eclipse-窗口-首选项-Java-已安装的JRE(对在当前开发环境中运行的java程序皆生效,也就是在eclipse中运行的java程序)编辑当前 ...
算法总结之欧拉函数&amp&semi;中国剩余定理
算法总结之欧拉函数&中国剩余定理 1.欧拉函数概念:在数论,对正整数n,欧拉函数是少于或等于n的数中与n互质的数的数目. 通式:φ(x)=x(1-1/p1)(1-1/p2)(1-1/p3)( ...
easyui常用控件及样式收藏
CSS类定义: div easyui-window window窗口样式属性如下: 1) modal:是否生成模态窗口.tru ...
ViewPager 详解（二）---详解四大函数
前言:上篇中我们讲解了如何快速实现了一个滑动页面,但问题在于,PageAdapter必须要重写的四个函数,它们都各有什么意义,在上节的函数内部为什么要这么实现,下面我们就结合Android的API说明 ...
javascript判断键盘按键
window.document.onkeydown = disableRefresh; function disableRefresh(evt){ evt = (evt) ? evt : window ...
框架篇：Spring+SpringMVC+hibernate整合开发
前言: 最近闲的蛋疼,搭个框架写成博客记录下来,拉通一下之前所学知识,顺带装一下逼. 话不多说,我们直接步入正题. 准备工作: 1/ IntelliJIDEA的安装配置:jdk/tomcat等..(本 ...
关于warning&colon; suggest parentheses around assignment used as truth value [-Wparentheses]|的解决方法
今天,在调试的时候一直出现warning: suggest parentheses around assignment used as truth value 代码如下: if(startTime== ...
IOT-SpringBoot-angular启动
1 D:\workspace_iot\iot-hub\src\main\angular cmd 启动 npm start 2 eclipse中启动springboot 3 local ...