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教育部為提昇全民資安素養與電腦防護能力，本部於101年9月5日至11月5日舉辦「全民資安素養自我評量」活動，請在活動期間內踴躍上網檢測資訊安全素養認知程度，並有機會參與抽獎，詳情請參閱活動網站（網址： https://isafe.moe.edu.tw/event
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url="https://www.twse.com.tw/exchangeReport/STOCK_DAY?response=json&date=20220330&stockNo=2330"
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先說明一下這是後知後覺的解答，所謂後知就是股票價格已知存在陣列(清單)，當然就要用迴圈練習，雙迴圈暴力解需時O(n**2)，當然不用，採python單一迴圈解答「最佳股票的買賣時機#LeetCode 121 Best Time to Buy and Sell Stock」，解...
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你會用C的算子sizeof?
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XOR prefixsum CPP python解Leetcode 1829 Maximum XOR for Each Query 其實這裡考慮的是元素個數為2^maximumBit的交換群，運算子為XOR。 [Python code請進]
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Python C++使用字串find迴圈速解Leetcode 2490 Circular Sentence C++ & Python都有字串的find，用法都類似 split=s.find(' ', split+1)能找到s從index=split+1開...
3 則留言:
#include <string>
#include <iostream>
using namespace std;
int main(int argc, char** argv) {
std::string horoscope[14];
horoscope[0] = "無此星座";
horoscope[1] = "摩羯座";
horoscope[2] = "水瓶座";
horoscope[3] = "雙魚座";
horoscope[4] = "白羊座";
horoscope[5] = "金牛座";
horoscope[6] = "雙子座";
horoscope[7] = "巨蟹座";
horoscope[8] = "獅子座";
horoscope[9] = "處女座";
horoscope[10] = "天秤座";
horoscope[11] = "天蠍座";
horoscope[12] = "蛇夫座";
horoscope[13] = "人馬座";
int n;
cout<<"Input n="
cin>>n;
cout<<horoscope[n]<<end1;
#include <string>
#include <iostream>
using namespace std;
int main(int argc, char** argv) {
string hor[14];
hor[0] = "無此星座";
hor[1] = "摩羯座";
hor[2] = "水瓶座";
hor[3] = "雙魚座";
hor[4] = "牡羊座";
hor[5] = "金牛座";
hor[6] = "雙子座";
hor[7] = "巨蟹座";
hor[8] = "獅子座";
hor[9] = "處女座";
hor[10] = "天秤座";
hor[11] = "天蠍座";
hor[12] = "蛇夫座";
hor[13] = "射手座";
int m = 0;
int d = 0;
cout << "輸入月份:";
cin >> m;
cout << "輸入日期:";
cin >> d;
int day = 0;
day = ((100 * m) + d);
int n = 0;
switch (m) {
case 1 :
if (day < 121) n = 13;
else n = 1;
break;
case 2 :
if (day < 217) n = 1;
else n = 2;
break;
case 3 :
if (day < 312) n = 2;
else n = 3;
break;
case 4 :
if (day < 419) n = 3;
else n = 4;
break;
case 5 :
if (day < 514) n = 4;
else n = 5;
break;
case 6 :
if (day < 622)n = 5;
else n = 6;
break;
case 7 :
if (day < 721) n = 6;
else n = 7;
break;
case 8 :
if (day < 811) n = 7;
else n = 8;
break;
case 9 :
if (day < 916) n = 8;
else n = 9;
break;
case 10 :
if (day < 1031) n = 9;
else n = 10;
break;
case 11 :
if (day < 1124) n = 10;
else if(day < 1130) n = 11;
else n = 12;
break;
case 12 :
if (day < 1218) n = 11;
else n = 12;
break;
default :
n = 0;
}
cout<<hor[n]<<endl;
return 0;
}
#include <string>
#include <iostream>
using namespace std;
int main(int argc, char** argv) {
string hor[14];
hor[0] = "無此星座";
hor[1] = "摩羯座";
hor[2] = "水瓶座";
hor[3] = "雙魚座";
hor[4] = "牡羊座";
hor[5] = "金牛座";
hor[6] = "雙子座";
hor[7] = "巨蟹座";
hor[8] = "獅子座";
hor[9] = "處女座";
hor[10] = "天秤座";
hor[11] = "天蠍座";
hor[12] = "蛇夫座";
hor[13] = "射手座";
int m = 0;
int d = 0;
cout << "輸入月份:";
cin >> m;
cout << "輸入日期:";
cin >> d;
int day = 0;
day = ((100 * m) + d);
int n = 0;
switch (m) {
case 1 :
if (day < 121) n = 13;
else n = 1;
break;
case 2 :
if (day < 217) n = 1;
else n = 2;
break;
case 3 :
if (day < 312) n = 2;
else n = 3;
break;
case 4 :
if (day < 419) n = 3;
else n = 4;
break;
case 5 :
if (day < 514) n = 4;
else n = 5;
break;
case 6 :
if (day < 622)n = 5;
else n = 6;
break;
case 7 :
if (day < 721) n = 6;
else n = 7;
break;
case 8 :
if (day < 811) n = 7;
else n = 8;
break;
case 9 :
if (day < 916) n = 8;
else n = 9;
break;
case 10 :
if (day < 1031) n = 9;
else n = 10;
break;
case 11 :
if (day < 1124) n = 10;
else if(day < 1130) n = 11;
else n = 12;
break;
case 12 :
if (day < 1218) n = 11;
else n = 12;
break;
default :
n = 0;
}
cout<<hor[n]<<endl;
return 0;
}
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