California Becomes The First State To Push Back School Start Times | TODAY

影片連結

pass:許可
mandate:指令、下令
drowsy:想睡的
behind:支持
upending:打亂
groggy:東倒西歪的
single mother:單親媽媽
too little too late:事情已經到了這地步，現在不管做甚麼動作，都來不及補救
second grader / fourth grader:二年級生/四年級生
Get off my lawn:滾開、不要多管閒事

井字遊戲

簡單的井字遊戲，純C硬刻~

程式碼:

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

char map[3][3] = {'\0'};
void initializeboard()
{
 int number = 1;
 for (int i = 0; i < 3; i++)
 {
  for (int j = 0; j < 3; j++)
   map[i][j] = ('0'+number++);
 }
}
void printborad()
{
 printf("-------------\n");
 printf("| %c | %c | %c |\n",map[0][0],map[0][1],map[0][2]);
 printf("----+---+---\n");
 printf("| %c | %c | %c |\n", map[1][0], map[1][1], map[1][2]);
 printf("----+---+----\n");
 printf("| %c | %c | %c |\n", map[2][0], map[2][1], map[2][2]);
 printf("-------------\n");
}
void playerturn()
{
 int cmd,correct_input=0;
 char mark='o';
 while (1)
 {
  printf("請選擇編號: ");
  scanf_s("%d", &cmd);
  if (cmd == 1 && map[0][0] == '1')
   map[0][0] = 'o', correct_input = 1;
  else if (cmd == 2 && map[0][1] == '2')
   map[0][1] = 'o', correct_input = 1;
  else if (cmd == 3 && map[0][2] == '3')
   map[0][2] = 'o', correct_input = 1;
  else if (cmd == 4 && map[1][0] == '4')
   map[1][0] = 'o', correct_input = 1;
  else if (cmd == 5 && map[1][1] == '5')
   map[1][1] = 'o', correct_input = 1;
  else if (cmd == 6 && map[1][2] == '6')
   map[1][2] = 'o', correct_input = 1;
  else if (cmd == 7 && map[2][0] == '7')
   map[2][0] = 'o', correct_input = 1;
  else if (cmd == 8 && map[2][1] == '8')
   map[2][1] = 'o', correct_input = 1;
  else if (cmd == 9 && map[2][2] == '9')
   map[2][2] = 'o', correct_input = 1;
  else
   printf("錯誤輸入，請再輸入一次!\n");

  if (correct_input)
   break;
 }
 
}
void computerturn()
{
 int x = rand() % 3, y = rand() % 3,i=0;
 while (map[x][y] == 'o' || map[x][y] == 'x')
 {
  x = rand() % 3, y = rand() % 3;
  i++;
  if (i >= 9)
   return;
 }
 map[x][y] = 'x';
 
}
int winorlose()
{
 // o win
 if (map[0][0] == 'o'&&map[0][1] == 'o'&&map[0][2] == 'o') return 1; //橫向
 if (map[1][0] == 'o'&&map[1][1] == 'o'&&map[1][2] == 'o') return 1;
 if (map[2][0] == 'o'&&map[2][1] == 'o'&&map[2][2] == 'o') return 1;
 if (map[0][0] == 'o'&&map[1][0] == 'o'&&map[2][0] == 'o') return 1; //縱向
 if (map[0][1] == 'o'&&map[1][1] == 'o'&&map[2][1] == 'o') return 1;
 if (map[0][2] == 'o'&&map[1][2] == 'o'&&map[2][2] == 'o') return 1;
 if (map[0][0] == 'o'&&map[1][1] == 'o'&&map[2][2] == 'o') return 1; //斜向
 if (map[0][2] == 'o'&&map[1][1] == 'o'&&map[2][0] == 'o') return 1;
 // x win
 if (map[0][0] == 'x'&&map[0][1] == 'x'&&map[0][2] == 'x') return 2;
 if (map[1][0] == 'x'&&map[1][1] == 'x'&&map[1][2] == 'x') return 2;
 if (map[2][0] == 'x'&&map[2][1] == 'x'&&map[2][2] == 'x') return 2;
 if (map[0][0] == 'x'&&map[1][0] == 'x'&&map[2][0] == 'x') return 2; //縱向
 if (map[0][1] == 'x'&&map[1][1] == 'x'&&map[2][1] == 'x') return 2;
 if (map[0][2] == 'x'&&map[1][2] == 'x'&&map[2][2] == 'x') return 2;
 if (map[0][0] == 'x'&&map[1][1] == 'x'&&map[2][2] == 'x') return 2; //斜向
 if (map[0][2] == 'x'&&map[1][1] == 'x'&&map[2][0] == 'x') return 2;

 int record = 0;
 for (int i = 0; i < 3; i++)
 {
  for (int j = 0; j < 3; j++)
  {
   if (map[i][j] == 'o' || map[i][j] == 'x')
    record++;
  }
 }
 if (record == 9) return 3;

 return 0;
}
int main()
{
 srand(time(NULL));
 printf("\n\n\n");
 printf("\t\t\t~~歡迎來玩 井字遊戲~~\n\n");
 printf("\t\t\t        <規則>\n");
 printf("\t\t\t1.玩家為先手，與電腦對決\n");
 printf("\t\t\t2.先連成一線者獲勝\n\n\n");
 system("pause");
 system("cls");
 initializeboard();

 int result = 0;
 while (1)
 {
  system("cls");
  printborad();
  playerturn();
  if ((result = winorlose()) != 0)
   break;
  computerturn();
  if ((result = winorlose()) != 0)
   break;
 }
 system("cls");
 printborad();
 if (result == 1)
  printf("你贏了~\n");
 else if (result == 2)
  printf("你輸了~\n");
 else if (result == 3)
  printf("平手~\n");
 system("pause");
 return 0;
}

What's it like to work at Google?

影片連結

無法觸及的夢幻之境啊~QQ

vibe:氛圍
gratifying:令人滿足的
forefront:最前線
leapfrog:越級提升
preconceived:先入為主的
push the boundaries:挑戰極限
reassuring:使安心
value:重視
show up:出席、到來

00147 - Dollars

解題思路:
零錢問題。我直接把0~300.00想成0~30000元去算，唯一要稍微注意的是小數在乘的時候會有誤差。

程式碼:

#include <stdio.h>
#define N 30000
long long int s[N+1]={0};
int C[11]={5,10,20,50,100,200,500,1000,2000,5000,10000};
int main()
{
    double n;
    s[0]=1;
    for(int i=0;i<11;i++)
    {
        for(int j=C[i];j<=N;j++)
            s[j]+=s[j-C[i]];
    }
    while(scanf("%lf",&n)&&n)
    {
        n+=1e-6;
        printf("%6.2lf",n);
        printf("%17lld\n",s[(int)(n*100)]);
    }
        
    return 0;
}

Australia's wildfires are still raging. Here's what you need to know.

影片連結

at any given time:隨時
lightning:閃電
inadvertently:不小心的
bushfire:山火、叢林大火
nature preserve:自然保留區
taking a (big) hit:受到打擊
fire crews :消防隊員

Wuhan Coronavirus: What You Should Know

影片連結

respiratory tract:呼吸道
pneumonia:肺炎
flu shot:流感疫苗
Influenza:流感
ventilator:呼吸器
etiquette:禮儀
hand gel:乾洗手

d140: On Sale

覺得自己的解法比較簡單，故紀錄之

解題思路:
大致上就是照著題目直觀去解，只是要注意1.誤差問題2.無條件捨去至第二位
1.誤差:由於小數在儲存時有誤差，運算時為避免錯誤要加上
2.無條件捨去:使用floor()函數(如求無條件進位則用ceil())

程式碼:

#include <stdio.h>
#include <math.h>

int main()
{
    
    double m;
    while(scanf("%lf",&m)!=EOF)
    {
        int check=0;
        if(m<100)
            check=1;
        if(m<=100)
            m*=0.9;
        else if(m<=500)
            m*=0.8;
        else
            m*=0.6;
        if(check)
            m+=8;
        m+=1e-6;
        printf("$%.2lf\n",floor(m*100)/100);
    }
    return 0;
}

c672: RGB ⇆ HEX

解題思路:
先了解RGB與HEX兩種分別是十進位與十六進位轉換，接著想辦法把三色分別取出即可。

程式碼:

#include <stdio.h>

int main()
{
    char s[12];
    while(gets(s)!=0)
    {
        if(s[0]=='#')
        {
            int r,g,b;
            sscanf(s,"#%02x%02x%02x",&r,&g,&b);
            printf("%d %d %d\n",r,g,b);
        }
        else
        {
            int r,g,b;
            sscanf(s,"%d %d %d",&r,&g,&b);
            printf("#%02X%02X%02X\n",r,g,b);
        }
    }
    return 0;
}

The Accidental Invention of the Best Snack Food Ever

影片連結

snarky:諷刺的
idiosyncratic:特別的
made their way to :(艱辛費時的)前往
mainstay:支柱
signature dish:招牌菜
hoity-toity:傲慢的；高不可攀的
aristocrat:貴族
scoff:嘲笑
soggy:浸水的
insanity:瘋狂；荒唐的行為
lost it:抓狂了
fire back:回擊
faux pas:失禮
diabolical:惡魔的
backfire:產生反效果
dig in:吃
thereafter:從那之後
took the world by storm:席捲世界

21點

很簡單的21點小遊戲。
無法不要牌繼續。

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int card[13] = { 0 };
int player_point = 0, computer_pount = 0, temp_record;

int get_card()
{
 int point = rand() % 13 + 1;//1~13
 while (card[point] >= 4)
 {
  point = rand() % 13 + 1;
 }
 card[point]++;

 return point;
}
void player_get_card()
{
 temp_record = get_card();
 player_point += temp_record;
 printf("你拿到一張牌: [%d]\n",temp_record);
 printf("你的總點數為: %d\n", player_point);
}
void computer_get_card()
{
 printf("電腦拿了一張牌\n");
 computer_pount += get_card();
}
void lose_or_win()
{
 if (player_point > 21 && computer_pount > 21)
  printf("兩方都輸啦!!");
 else if (player_point > 21)
  printf("你輸啦!!");
 else if (computer_pount > 21)
  printf("你贏啦!!");
 else if (player_point > computer_pount)
  printf("你贏啦!!");
 else if (player_point < computer_pount)
  printf("你輸啦!!");
 else if (player_point == computer_pount)
  printf("平手啦!!");
}
int main()
{
 srand(time(NULL));
 char cmd;

 printf("\n\n");
 printf("~~歡迎來玩 21點 小遊戲~~\n\n");
 system("pause");
 system("cls");

 player_get_card();
 computer_get_card();


 while (1)
 {
  printf("-----------------------");
  printf("\n\n");
  printf("請問要繼續嗎? (y/n)\n");
  printf("y: 繼續 n:結束此局\n");
  scanf_s("%c", &cmd);
  getchar();
  if (cmd == 'n' || cmd == 'N')
   break;
  printf("-----------------------");
  printf("\n");

  player_get_card();
  computer_get_card();

  if (player_point > 21 || computer_pount > 21)
   break;
 }
 printf("-----------------------\n\n");
 printf("你的點數: %d\n", player_point);
 printf("電腦的點數: %d\n", computer_pount);
 printf("\n");
 
 lose_or_win();
 system("pause");
 return 0;
}

早餐生成器

有選擇障礙症，因此產出的簡單小程式

#include <stdio.h>
#include <time.h>
#include <stdlib.h>
void breakfast()
{
 int food = rand() % 8;
 printf("\n");
 char  menu[10][10] = { "包子","三明治","香腸炒蛋","香腸","炒蛋","蛋餅","蔥油餅","絕食" };
 printf("就決定是你了 ──  %s !!\n\n", menu[food]);
}
void lunch()
{
 int food = rand() % 7;
 printf("\n");
 char menu[20][20] = { "炒飯","池上便當","燒賣","水餃","鍋貼","便當","媽媽愛心午餐" };
 printf("那就吃  %s  好了~\n\n", menu[food]);
}
void dinner()
{
 int food = rand() % 8;
 printf("\n");
 char menu[20][20] = { "火鍋","壽司","麥當勞","摩斯","吉野家","炒飯","漢堡王","披薩" };
 printf("吃  %s  感覺很不錯!\n\n", menu[food]);
}
int main()
{
 int command;
 system("color f0");
 srand(time(NULL));
 printf("================\n");
 printf("歡迎來到 菜單生成器\n");
 printf("================\n");
 printf("\n\n");
 while (1)
 {
  char c;
  printf("1) 早餐\n");
  printf("2) 午餐\n");
  printf("3) 晚餐\n");
  printf("請輸入選項: ");
  scanf_s("%d", &command);
  getchar();
  if (command == 1)
   breakfast();
  else if (command == 2)
   lunch();
  else if (command == 3)
   dinner();
  else
   printf("請好好輸入好嗎=口=\n");
  printf("\n");
  while (1)
  {
   printf("還要繼續嗎? y=繼續   n=結束\n");
   scanf_s("%c", &c);
   getchar();
   if (c == 'n' || c == 'N' || c == 'y' || c == 'Y')
    break;
   else
    printf("請好好輸入好嗎=口=\n\n");
   printf("\n");
  }
  if (c == 'n' || c == 'N')
   break;
  printf("\n");
 }
 printf("~~~~謝謝你的光臨~~~~\n");
 system("pause");
 return 0;
}

Chernobyl (2019) | Official Trailer | HBO

影片連結

不看字幕就看不懂....程度太差QQ

justice was done:伸張正義
just:公正的、正義的
sane:理智的
on fire:起火
uranium:鈾
firing:發射
contain:遏止
misinformation:誤傳
madness:瘋狂的行為

亡靈代言人(SPEAKER FOR THE DEAD) 單字筆記

把整本小說中一再出現的單字紀錄一下。

10098 - Generating Fast, Sorted Permutation

這題zerojudge上測資有問題，丟uva就過了，傻眼orz
其實對這種字串排序還是不怎麼熟，是把工具套一套解出來的。

步驟:
1.對字串進行排序
2.next_permutation

程式碼:

#include <iostream>
#include <algorithm>
using namespace std;

int main()
{
    int n;
    cin>>n;
    getchar();
    while(n--)
    {
        string s;
        getline(cin,s);
        int len=s.size();
        sort(s.begin(),s.end());
        do{
            cout<<s<<endl;
        }while(next_permutation(s.begin(),s.end()));
        cout<<endl;
    }
    return 0;
}

參考:https://stackoverflow.com/questions/17006808/find-all-permutations-of-a-string-in-c

next_permutation是字典序較大的放右邊，而與之相反的則是prev_permutation。

相關題目:d703: SOS ~~~

Opinion | Trump's racist tweets will not age well

影片連結

Democrat:民主黨員
congresswoman:女眾議員
inept:無能的
viciously:邪惡的
infest:(動物、害蟲)大批出沒
tweet:在推特上發文
common cause:共同追求的目標
duly:合適的
Israel:以色列

Ivan Joseph | 5 Keys to self confidence (Key Points Talk)

影片連結

repetition:重複
ounce:盎司
bail:放棄
self talk:自我對話
score a goal :進球得分
demoralized:士氣低落、意志消沉

Paolo Cardini: Forget multitasking, try monotasking

影片連結

supertasker:能同時完成多件事的人
front cover:手機殼(但網路查的結果都是:封面，可能這裡只是單純想表達「把前面蓋住的東西」)
downgrade:降級
Venice:威尼斯
How beautiful it is to lose ourselves in these little streets:這些小街道是如此美麗的令人忘我

Arthur Benjamin: Teach statistics before calculus!

影片連結

Don't get me wrong:別誤會我的意思
day-to-day:日常的
mess:混亂局面
analog:類比
discrete:離散的
standard deviation:標準差

How to tie your shoes | Terry Moore

影片連結

worldly:善於處世的
savvy:有領悟力
ludicrous:荒唐的
laces:鞋帶
nail:成功學會
strand:繩
orient:使朝向
long axis:長軸
bow:結
start over:重來
cord:繩、線
transverse:縱向的
in keeping with:呼應、與...一致
Live long and prosper:生生不息，繁榮昌盛(源自星際大戰)

質數

質數篩法

#define N 100
int isprime[N];
int prime[N],pnum;
void sieve(int n)
{
    pnum=0;
    for(int i=0;i<n;i++)
        isprime[i]=1;
    isprime[0]=isprime[1]=0;
    int sqn=sqrt(n);
    for(int i=2;i<=sqn;i++)
    {
        if(isprime[i])
        {
            prime[pnum++]=i;
            for(int j=i*i;j<n;j+=i)
                isprime[j]=0;
        }
    }
    for(int i=sqn+1;i<=n;i++)
    {
        if(isprime[i])
            prime[pnum++]=i;
    }
}

其他

int isprime[N]={1,1};
    int sqn=sqrt(N);
    for(int i=2;i<=sqn;i++)
    {
        if(!isprime[i])
        {
            for(int j=i*i;j<N;j+=i)
                isprime[j]=1;
        }
    }

int Prime[100], Pt;
void Sieve() {
 Pt = 0;
 int i, j, mark[100] = {0};
 Prime[Pt++] = 2;
 for(i = 3; i < 100; i += 2) 
 {
  if(mark[i] == 0) 
  {
   Prime[Pt++] = i;
   for(j = 3; i*j < 100; j += 2)
    mark[i*j] = 1;
  }
 }
}

質因數

#include <stdio.h>
#include <math.h>
#define M 1000
#define N 10000
int fact[M],count[M],fnum;
int prime[N],isprime[N],pnum;
void sieve()
{
    int sqn=sqrt(N);
    isprime[0]=isprime[1]=1;
    for(int i=2;i<N;i++)
    {
        if(!isprime[i])
        {
            for(int j=i+i;j<=N;j+=i)
                isprime[j]=1;
        }
    }
    pnum=0;
    for(int i=0;i<N;i++)
    {
        if(!isprime[i])
            prime[pnum++]=i;
    }
}
void factorization(int n)
{
    fnum=0;
    for(int i=0;i<pnum&&n>1;i++)
    {
        if(n%prime[i]==0)
        {
            fact[fnum]=i;
            count[fnum]=0;
            fnum++;
            while(n%prime[i]==0)
            {
                count[fnum-1]++;
                n/=prime[i];
            }
        }
    }
}
int main()
{
    sieve();
    factorization(120);
    printf("%d = \n",120);
    for(int i=0;i<fnum;i++)
        printf("%d^%d\n",prime[fact[i]],count[i]);
    return 0;
}

8 secrets of success

影片連結

make something of one's life:有所成就
pass it on to :傳達給
what makes (one) tick:促使某人做某事的原因
workafrolic:作者自創字。對比工作狂(workaholic)，字根frolic有:「快樂做某事」的意思，因此字幕翻譯成工作玩家。
put your nose down in something:專注某事，實際去做(不確定意思)