codejam_2014_qualification

Problem A. Magic Trick

#include <iostream>#include <algorithm>#include <vector>using namespace std;void read(vector<int> &vec) {  int card;  for (int i = 0; i < 16; ++i) {    cin >> card;    vec.push_back(card);  }  return;}int main() {  int T, row1, row2, card;  cin >> T;  for (int c = 1; T--; c++) {    vector<int> arrange1, arrange2;    cin >> row1;    read(arrange1);    cin >> row2;    read(arrange2);    int cnt = 0;    auto iter1 = arrange2.begin() + (row2 - 1) * 4, iter2 = iter1 + 4;    for (int i = 0; i < 4; ++i) {      if (find(iter1, iter2, arrange1[(row1 - 1) * 4 + i]) != iter2) {        ++cnt;        card = arrange1[(row1 - 1) * 4 + i];      }    }    if (cnt == 1)      cout << "Case #" << c << ": " << card << endl;    else if (cnt > 1)      cout << "Case #" << c << ": " << "Bad magician!" << endl;    else      cout << "Case #" << c << ": " << "Volunteer cheated!" << endl;  }  return 0;}

Problem B. Cookie Clicker Alpha

    迭代，每次够C的时候进行判断，是买新fram达到X速度快还是不买速度快，如果某次判断不买farm则以后也不需要买。

#include <stdio.h>int main() {  double c, f, x;  int T;  scanf("%d", &T);  for (int cnt = 1; T--; ++cnt) {    scanf("%lf %lf %lf", &c, &f, &x);    double spd = 2, time_sum = 0, time_y, time_n;    while (x > c) {      time_sum += c / spd;      time_y = x / (spd + f);      time_n = (x - c) / spd;      if (time_y <= time_n) {        spd += f;      } else {        x -= c;        break;      }    }    time_sum += x / spd;    printf("Case #%d: %0.7lf\n", cnt, time_sum);  }  return 0;}

Problem D. Deceitful War

    主要是看明白体意，模拟两种游戏策略即可。

    在实现的时候用set比较浪费，因为给每种策略的函数传参数时候都要传值，因为函数中会对set进行修改，比较好的方法是用vector，然后排序或者用数组来记录烧掉的blocks，这样可以传引用。

    另外还要注意，由于set的底层是红黑树，非线性结构，因此iterator没有operator-（不能像vector等进行iterator的+ - 操作），但是有++和--。

#include <stdio.h>#include <set>using namespace std;int Deceitful(set<double> naomi, set<double> ken, int n) {  int ds = 0;  while (n--) {    auto iter_n1 = naomi.begin();    auto iter_k1 = ken.begin(), iter_k2 = ken.end();    if (*iter_n1 > *iter_k1) {      ++ds;      naomi.erase(iter_n1), ken.erase(iter_k1);    } else {      naomi.erase(iter_n1), ken.erase(--ken.end());    }  }  return ds;}int War(set<double> naomi, set<double> ken, int n) {  int ws = 0;  // 1:min 2:max  while (n--) {    auto iter_n1 = naomi.begin();    auto iter_k1 = ken.begin(), iter_k2 = ken.end();    for ( ; iter_k1 != iter_k2 && *iter_n1 > *iter_k1; ++iter_k1) {}    if (iter_k1 == iter_k2) {      // cannot use naomi.end() - 1, set is no operator -      ken.erase(ken.begin()), naomi.erase(--(naomi.end()));      ++ws;    } else {      ken.erase(iter_k1), naomi.erase(iter_n1);    }  }  return ws;}void Read(set<double> &st, int n) {  double block;  for (int i = 0; i < n; ++i) {    scanf("%lf", &block);    st.insert(block);  }  return;}int main() {  int T, N;   scanf("%d", &T);  for (int c = 1; T--; ++c) {    set<double> naomi, ken;    scanf("%d", &N);    Read(naomi, N), Read(ken, N);    // must pass value, too cost! use vector is more better.    int ds = Deceitful(naomi, ken, N);    int ws = War(naomi, ken, N);    printf ("Case #%d: %d %d\n", c, ds, ws);  }  return 0;}