********************** 第一題 d061: 曾祖母的慶生會 **********************
#include <iostream>
using namespace std;
int main () {
int y;
cin >> y;
cout << 97 - y - (y < 0) << endl;
}
********************** 第二題 b202: A. 優惠方案 **********************
#include <iostream>
using namespace std;
int main () {
int k, p1, p2, p3;
cin >> k;
while (k--) {
cin >> p1 >> p2 >> p3;
if (p1 != p2 && p2 != p3 && p1 != p3)
cout << "YES\n";
else
cout << "NO\n";
}
}
********************** 第三題 d669: 11677 - Alarm Clock **********************
#include <iostream>
using namespace std;
int main () {
int h1, m1, h2, m2;
while (cin >> h1 >> m1 >> h2 >> m2, h1 + m1 + h2 + m2) {
cout << ((h2*60 + m2) - (h1*60 + m1) + 1440) % 1440 << endl;
}
}
********************** 第四題 d980: 11479 - Is this the easiest problem? **********************
#include <iostream>
using namespace std;
int main () {
long long T, x, a, b, c;
cin >> T;
for (x=1; x<=T; x++) {
cin >> a >> b >> c;
if (a <= 0 || b <= 0 || c <= 0 || a+b <= c || a+c <= b || b+c <= a)
cout << "Case " << x << ": Invalid\n";
else if (a == b && b == c)
cout << "Case " << x << ": Equilateral\n";
else if (a == b || b == c || c == a)
cout << "Case " << x << ": Isosceles\n";
else
cout << "Case " << x << ": Scalene\n";
}
}
********************** 第五題 b226: E. 鋪地磚 **********************
#include <iostream>
using namespace std;
int main () {
int L, W, x, y;
while (cin >> L >> W >> x >> y, L) {
cout << ((L%x || W%y) && (L%y || W%x) ? -1 : L * W / x / y) << endl;
} //若直擺不行 橫擺也不行
}
********************** 第六題 b080: A. 畢氏定理 **********************
#include <iostream>
#include <cmath>
using namespace std;
int main () {
int a, b, c, c2; //c2--c 平方
while (cin >> a >> b, a) {
c2 = a*a + b*b; //假設c 為斜邊
c = (int)sqrt((double)c2);
if (c2 == c*c)
cout << c << endl;
else {
if (a < b) swap (a, b); //確定a > b
c2 = a*a - b*b; //假設a 為斜邊
c = (int)sqrt((double)c2);
if (c2 == c*c)
cout << c << endl;
else
cout << "Wrong\n";
}
}
}
********************** 第七題 a132: 10931 - Parity **********************
#include <iostream>
#include <string>
using namespace std;
int main () {
int n, i, s;
string b, d="01";
while (cin >> n, n) {
s = 0;
b = "";
while (n) {
s += n % 2;
b = d[n % 2] + b;
n /= 2;
}
cout << "The parity of " << b << " is " << s << " (mod 2).\n";
}
}
********************** 第八題 b225: D. 棒球練習 **********************
#include <iostream>
using namespace std;
int main () {
int N;
double x1, y1, x2, y2, x3, y3;
cin >> N;
while (N--) {
cin >> x1 >> y1 >> x2 >> y2 >> x3 >> y3;
cout << ((x2-x1)*(y3-y1) == (x3-x1)*(y2-y1) ? "NO\n" : "YES\n");
} //若斜率相等則不成一個三角形
}
********************** 第九題 b072: D. 打不倒的空氣人 **********************
#include <iostream>
#include <string>
using namespace std;
int main () {
string s, w, pw;
int n, m, A;
while (cin >> n >> m, n) {
s = " "; //用一個空白佔住s[0]
while (n--) {
cin >> w; //w(ord)--英文單字
s += w; //s = s + w; 會超時
}
pw = ""; //p(ass)w(ord)--密碼
while (m--) {
cin >> A;
pw += s[A];
}
cout << pw << endl;
}
}
********************** 第十題 a130: 12015 - Google is Feeling Lucky **********************
#include <iostream>
#include <string>
using namespace std;
int main () {
int i, mxv, v[10], t, tc=1;
string u[10]; //u(rl)--網址
cin >> t;
while (t--) {
mxv = 0; //max v--最大相關度
for (i=0; i<10; i++) {
cin >> u[i] >> v[i];
mxv = max (mxv, v[i]); //求v[i] 最大值
}
cout << "Case #" << tc++ << ":\n";
for (i=0; i<10; i++)
if (v[i] == mxv) //若相關度最高
cout << u[i] << endl;
}
}