2010 校慶盃乙組參考解答

//11677 - Alarm Clock (by Snail)
#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;
     }
}

//10921 - Find the Telephone (by Snail)
#include <iostream>
#include <string>
using namespace std;

int main () {
     int i;
     string s, x1="-01ABCDEFGHIJKLMNOPQRSTUVWXYZ", x2="-0122233344455566677778889999";
     while (cin >> s) {
           for (i=0; i<s.size(); i++)
                s[i] = x2[x1.find(s[i])];
           cout << s << endl;
     }
}

//11716 - Digital Fortress (By Snail)
#include <iostream>
#include <string>
#include <cmath>
using namespace std;

int main () {
     string s;
     int t, i, j, r;
     cin >> t;
     getline (cin, s);                             // 讀掉 t 後面的換行
     while (t--) {
           getline (cin, s);
           r = (int) sqrt ((double)s.size());             // r(oot)-- 平方根
           if (r*r == s.size()) {                         // 完全平方數
                for (i=0; i<r; i++)
                     for (j=i; j<(int)s.size(); j+=r)
                           cout << s[j];
                cout << endl;
           } else
                cout << "INVALID\n";
     }
}

//10922 - 2 the 9s (by Snail)
#include <iostream>
#include <string>
using namespace std;

int main () {
     string n;
     int i, s, t, d;
     while (cin >> n, n != "0") {
          s = 0;
          for (i=0; i<(int)n.size(); i++)          // 求所有位數的數字和
                s += n[i] - '0';
           if (s % 9)                                     // 不是 9 的倍數
                cout << n << " is not a multiple of 9.\n";
           else {
                for (d=1; s>9; d++) {                     // d(egree)
                     for (t=s, s=0; t; t/=10)             // 求 t 的數字和
                           s += t % 10;
                }
                cout << n << " is a multiple of 9 and has 9-degree " << d << ".\n";
           }
     }
}

//11608 - No Problem (by Snail) -- 無陣列暴力版
#include <iostream>
using namespace std;

int main () {
     int s, r, n1, n2, n3, n4, n5, n6, n7, n8, n9, n10, n11, n12, x=1;
     while (cin >> s, s >= 0) {
           cin >> n1 >> n2 >> n3 >> n4 >> n5 >> n6 >> n7 >> n8 >> n9 >> n10 >> n11 >> n12;

           cout << "Case " << x++ << ":\n";

           cin >> r;                                       // 一月
           if (s >= r) {                                   // 題目足夠
                cout << "No problem! :D\n";
                s -= r;                                    // 減掉用掉的題目
           } else
                cout << "No problem. :(\n";
           s += n1;                                        // 加上本月所出題目

           cin >> r;                                       // 二月
           if (s >= r) {
                cout << "No problem! :D\n";
                s -= r;
           } else
                cout << "No problem. :(\n";
           s += n2;

           // 重覆恕刪

           cin >> r;                                       // 十二月
           if (s >= r) {
                cout << "No problem! :D\n";
                s -= r;
           } else
                cout << "No problem. :(\n";
           s += n12;
     }
}

//10192 - Vacation (by Snail)
#include <iostream>
#include <string>
using namespace std;

int main() {
     int C[101][101], tc=1, i, j;
     string X, Y;
     while (getline(cin, X), X != "#") {
           getline(cin, Y);
           for (i=0; i<=X.size(); i++) C[i][0] = 0;
           for (j=1; j<=Y.size(); j++) C[0][j] = 0;
           for (i=1; i<=X.size(); i++)
                for (j=1; j<=Y.size(); j++)
                     if (X[i-1]==Y[j-1])
                           C[i][j] = C[i-1][j-1] + 1;
                     else
                           C[i][j] = max(C[i][j-1], C[i-1][j]);
           cout << "Case #" << tc++ << ": you can visit at most " << C[X.size()][Y.size()] << " cities.\n";
     }
}