102-2 資訊第一次小考參考解答

*************** 第一題 ***************

//a040: 阿姆斯壯數 by Snail

#include <iostream>

using namespace std;

int main () {

    int i, j, a, b, arm[21]={1,2,3,4,5,6,7,8,9,153,370,371,407,1634,8208,9474,54748,92727,93084,548834,1000001};

    //int k, r, n, x, e, d[7], narm=0;  //求陣列中的值

    //for (i=1; i<=1000000; i++) {

    //  r = i;

    //  for (e=0; r; e++) {

    //      d[e] = r % 10;

    //      r /= 10;

    //  }

    //  n = 0;

    //  for (j=0; j<e; j++) {

    //      x = 1;

    //      for (k=0; k<e; k++)

    //          x *= d[j];

    //      n += x;

    //  }

    //  if (n == i)

    //      arm[narm++] = n;

    //}

    //arm[narm] = 1000001;              //陣列尾端記號

    while (cin >> a >> b) {

        for (i=0; arm[i]<a; i++);

        for (j=i; arm[j]<=b; j++)

            cout << arm[j] << ' ';

        cout << (i==j ? "none\n" : "\n");

    }

}

*************** 第二題 ***************

//11608 - No Problem (by Snail)

#include<iostream>

using namespace std;

int main (){

    int s,i,a[100],b,t=1;

    while(cin>>s, s>=0){

        for(i=0; i<12; i++)

            cin>>a[i];

        cout << "Case " << t++ << ":" << endl;

        for(i=0; i<12; i++){

            cin>>b;

            if(b>s)

                cout << "No problem. :(" << endl;

            else {

                cout << "No problem! :D" << endl;

                s -= b;

            }

            s += a[i];

        }

    }

}

*************** 第三題 ***************

//d212. 東東爬階梯 by Snail

#include <iostream>

using namespace std;

int main () {

    unsigned long long s[93] = {1, 1};

    int i, n;

    for (i=2; i<=92; i++)

        s[i] = s[i-1] + s[i-2];

    i = 0;

    while (cin >> n)

        i=max(i,n),cout << s[n] << endl;

}

*************** 第四題 ***************

//382 - Perfection (by Snail)

#include <iostream>

#include <string>

#include <iomanip>

using namespace std;

int main () {

    int s, n, f;

    string output[3] = {"DEFICIENT", "PERFECT", "ABUNDANT"};

    cout << "PERFECTION OUTPUT\n";

    while (cin >> n, n) {

        s = n>1;

        for (f=2; f*f<n; f++)

            if (n % f == 0)

                s += f + n/f;

        if (f*f == n)

            s += f;

        cout << setw(5) << n << "  " << output[(s>=n)+(s>n)] << endl;

    }

    cout << "END OF OUTPUT\n";

}

*************** 第五題 ***************

//10924 - Prime Words (by Snail)

#include <iostream>

#include <string>

using namespace std;

int main () {

    int i, n, n2;

    string s;

    while (cin >> s) {

        n = 0;

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

            n += (s[i] < 'a' ? s[i] - 38 : s[i] - 96);

        for (i=2; n%i && i*i<=n; i++);                  //求第一個 <= 平方根的因數

        cout << (i*i > n ? "It is a prime word.\n" : "It is not a prime word.\n");

    }

}

*************** 第六題 ***************

//D. 跑跑卡丁車 -- 2009 NPSC 高中組決賽 by Snail

#include <iostream>

#include <string>

#include <algorithm>

using namespace std;

struct racer {

    string n, t;

    int sn;

} r[100000];

bool timelt (racer a, racer b) {return a.t < b.t;}

bool snlt (racer a, racer b) {return a.sn < b.sn;}

int main () {

    int n, i, k, h, m;

    char ch;

    while (cin >> n, n) {

        for (i=0; i<n; i++) {

            cin >> r[i].n >> r[i].t;

            r[i].sn = i;                        //s(erial) n(umber)--序號

        }

        sort (r, r+n, timelt);                  //依時間由小到大排序

        for (k=n/3; k<n && r[k].t==r[k-1].t; k++);//同分增額錄取

        sort (r, r+k, snlt);                    //依序號由小到大排序

        cout << "LIST START\n";

        for (i=0; i<k; i++)

            cout << r[i].name << endl;

        cout << "LIST END\n";

    }

}

//D. 跑跑卡丁車 -- 2009 NPSC 高中組決賽 by kevin704300 (邱玉全)

#include <iostream>

#include <string>

#include <algorithm>

using namespace std;

string a[100000], b[100000], c[100000];

int main () {

    int n, i, j, m;

    while (cin >> n, n) {

        cout << "LIST START\n";

        for (i=0; i<n; i++)

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

        sort(c, c+n);

        m = n/3 - 1;

        for (i=0; i<n; i++)

            if (b[i] <= c[m])

                cout << a[i] << endl;

        cout << "LIST END\n";

    }

}