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2016ccf認(rèn)證考試試題

　　CDMA認(rèn)證論壇，即CCF是一個由CDMA運(yùn)營商和制造商共同建立和維護(hù)的全球性非營利組織。該組織致力于建立和規(guī)范CDMA終端產(chǎn)品的全球統(tǒng)一認(rèn)證和流程，促進(jìn)CDMA終端產(chǎn)品互操作性和一致性，推動CDMA產(chǎn)業(yè)的發(fā)展。2009年9月10日，CTIA和CDG組織將其CDMA產(chǎn)品認(rèn)證部分統(tǒng)一合并入CCF認(rèn)證，自此，CCF認(rèn)證成為全球CDMA終端官方統(tǒng)一認(rèn)證。

2016ccf認(rèn)證考試試題

　　1. 出現(xiàn)次數(shù)最多的數(shù)

　　#include #i

　　nclude #include #include #include #include #includeusing namespace std; int main() { int n; cin >> n; map f; for (int i = 0; i < n; i++) { int t; cin >> t; f[t]++; } int ans, m = 0; for (map::iterator it = f.begin(); it != f.end(); it++) { if (it->second > m) { m = it->second; ans = it->first; }

　　} cout << ans << endl; return 0; }

　　2. ISBN 號碼

　　#include #include #include #include #include #include #includeusing namespace std; int a[10]; int main() { string s; cin >> s; a[0] = s[0] - '0'; a[1] = s[2] - '0'; a[2] = s[3] - '0'; a[3] = s[4] - '0'; a[4] = s[6] - '0'; a[5] = s[7] - '0'; a[6] = s[8] - '0'; a[7] = s[9] - '0'; a[8] = s[10] - '0'; a[9] = s[12] - '0'; int sum = 0; for (int i = 0, j = 1; i < 9; i++, j++) { sum += a[i] * j; } int code = sum % 11; char c = code == 10 ? 'X' : '0' + code; if (s[12] == c) {

　　cout << "Right" << endl; }else { s[12] = c; cout << s << endl; } return 0; }

　　3.最大的矩形

　　#include #include #include #include #include #include using namespace std; int main() { int n; vector a; cin >> n; for (int i = 0; i < n; i++) { int x; cin >> x; a.push_back(x); } int ans = 0; for (int i = 0; i < n; i++) { int h = a[i]; for (int j = i; j < n; j++) { if (a[j] < h) h = a[j]; int s = (j - i + 1) * h; if (ans < s) ans = s; }

　　} cout << ans << endl; return 0; }

　　4.有趣的數(shù)

　　#include #include #include #include #include #include #include using namespace std; long long f[2000][3][2]; // f[seq_k to place][0: to place 0 , 1: ethier 0 or 1, 2 : must be 1][3 is placed ? 1 : 0] int dp(int n, int p1, int p3) { long long &now = f[n][p1][p3]; if (now != -1) return now; if (n == 0) { if (p1 == 2 && p3 == 1) { now = 1; }else { now = 0; } return now; } now = 0; if (p1 == 0) { now += dp(n-1, 1, p3); // go 0 }else if (p1 == 1) { now += dp(n-1, 1, p3); // go 0

　　now += dp(n-1, 2, p3); // go 1 }else // p1 == 2 { now += dp(n-1, 2, p3); // go 1 } if (p3 == 0) { now += dp(n-1, p1, p3); // go 2; now += dp(n-1, p1, 1); // go 3; }else { now += dp(n-1, p1, 1); // go 3; } now %= 1000000007; } int main() { int n; cin >> n; memset(f, -1, sizeof(f)); int ans = dp(n - 1, 0, 0); // seq[n] is 2 cout << ans << endl; return 0; }

　　5.I’m stuck!

　　#include #include #include #include #include #include #include using namespace std; // class Move {

　　public: virtual bool CanMove(char from, char to, int dx, int dy) = 0; }; class ForwardMove : public Move { public: virtual bool CanMove(char from, char to, int dx, int dy) { if (to == '#') return false; switch (from) { case '+' : case 'S' : case 'T' : return true; break; case '-' : return dy != 0; break; case '|' : return dx != 0; break; case '.' : return dx == 1; break; } return false; } }; class BackwardMove : public Move { public: virtual bool CanMove(char from, char to, int dx, int dy) { if (to == '#') return false; switch (to) { case '+' : case 'S' : case 'T' : return true; break; case '-' : return dy != 0; break; case '|' : return dx != 0; break; case '.' : return dx == -1; break; } return false; } }; char s[100][100]; typedef bool ARR[100][100]; ARR bs, bt; int sx, sy, tx, ty;

　　int d[4][2] = {{-1, 0},{1, 0},{0, 1},{0, -1}}; void Bfs(ARR b, Move *move, int x, int y) { if (b[x][y]) return; b[x][y] = true; for (int o = 0; o < 4; o++) { int dx = d[o][0]; int dy = d[o][1]; int xx = x + dx; int yy = y + dy; if (move->CanMove(s[x][y], s[xx][yy], dx, dy)) { Bfs(b, move, xx, yy); } } } int n, m; int main() { cin >> n >> m; for (int i = 0; i <= n + 1; i++) for (int j = 0; j <= m + 1; j++) s[i][j] = '#'; for (int i = 1; i <= n; i++) cin >> s[i]+1; for (int i = 0; i <= n + 1; i++) s[i][m + 1] = '#';

　　for (int i = 0; i <= n + 1; i++) { for (int j = 0; j <= m + 1; j++) { if (s[i][j] == 'S') { sx = i;

　　sy = j; } if (s[i][j] == 'T') { tx = i; ty = j; } } } Bfs(bs, new ForwardMove(), sx, sy); Bfs(bt, new BackwardMove(), tx, ty); int ans = 0; for (int i = 0; i <= n + 1; i++) { for (int j = 0; j <= m + 1; j++) { if (bs[i][j] && ! bt[i][j]) ans ++; } } if (bs[tx][ty] == false) cout << "I'm stuck!" << endl; else cout << ans << endl; return 0; }

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ccf認(rèn)證考試試題

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