vmware以及schlumberger题解

先是vmare的：具体的题目我就不描述了。

1. 贪吃的小明。直接数个数，统计个数，就可以完成。使用map，应该输入implement这一类，我认为很简单，但是我只过了33%。

 /*
 ID: y1197771
 PROG: test
 LANG: C++
 */
 #include<bits/stdc++.h>
 #define pb push_back
 #define FOR(i, n) for (int i = 0; i < (int)n; ++i)
 #define dbg(x) cout << #x << " at line " << __LINE__ << " is: " << x << endl
 typedef long long ll;
 using namespace std;
 typedef pair<int, int> pii;
 const int maxn = 1e3 + ;
 string s[] = {"ONE", "TWO", "THREE", "FOUR", "FIVE", "SIX" };
 void solve() {
     int n, a, b, c;
     string s1, s2;
     a = b = c = ;
     map<string, int> m;
     cin >> n;
     for (int i = ; i < ; i++) m[s[i] ] = i;
     for (int i = ; i < n; i++) {
         cin >> s1 >> s2;
         if(m[s1] > m[s2]) a++;
         else if(m[s1] < m[s2]) b++;
         else c++;
     }
     cout << a << " " << b << " " << c << endl;
     if(b < a) {
         cout << "MingMing Win" << endl;
     } else {
         cout << "LiangLiang Win" << endl;
     }

 }
 int main() {
     //freopen("test.in", "r", stdin);
     //freopen("test.out", "w", stdout);
     solve();
     return ;
 }

2. 贾昆的谍报计划，可以从任意点开始，最长下降路径，更leetcode https://leetcode.com/problems/longest-increasing-path-in-a-matrix/这道题一致吧， 我认为是一样，因为上升和下降是完全一致的。这题我也没100%ac，但是我写的在leetcode可以ac啊，我不知道什么情况。我好想知道怎么回事了，set判重导致的错误，换priority_queue就可以。

 /*
 ID: y1197771
 PROG: test
 LANG: C++
 */
 #include<bits/stdc++.h>
 #define pb push_back
 #define FOR(i, n) for (int i = 0; i < (int)n; ++i)
 #define dbg(x) cout << #x << " at line " << __LINE__ << " is: " << x << endl
 typedef long long ll;
 using namespace std;
 typedef pair<int, int> pii;
 const int maxn = 1e3 + ;
 int a[][];
 int cnt[][];
 struct node {
     int v, x, y;
     bool operator<(const node & t) const {
         return v < t.v;
     }
 };
 int dx[] = {, , , -};
 int dy[] = {, , -, };
 void solve() {
     set<node> se;
     int n, m, x, y; cin >> n >> m;
     for (int i = ; i < n; i++) {
         for (int j = ; j < m; j++) {
             cin >> x; a[i][j] = x;
             se.insert({x, i, j});
             cnt[i][j] = ;
         }
     }
     int res = ;
     while(!se.empty()) {
             int p = se.begin()->v;
         x = se.begin()->x; y = se.begin()->y;
         //cout << p << " " << x << " " << y << " " << res << endl;
         res = max(res, cnt[x][y]);
         se.erase(se.begin());
         for (int i = ; i < ; i++) {
             int cx = x + dx[i], cy = y + dy[i];
             if(cx <  || cx >= n || cy <  || cy >= m) continue;
             if(a[cx][cy] <= p) continue;
             cnt[cx][cy] =  max(cnt[x][y] + , cnt[cx][cy]);
         }
     }
     cout << res << endl;
 }
 int main() {
     freopen("test.in", "r", stdin);
     //freopen("test.out", "w", stdout);
     solve();
     return ;
 }

3.旗帜计数，这道题比较变态，很难，后来查了一下，是cf原题，转移方程+快速幂，转移方程不好搞，链接在这里http://codeforces.com/problemset/problem/93/D注意，这题是div1的第4道题，当时ac的不到100人左右。

schlimberger

1. mumuchacha的珍珠手链，求固定窗口内最大值，直接滑动窗口O（1）的转移，很简单，或者是前缀和，代码难度比较低。直接过。

 /*
 ID: y1197771
 PROG: test
 LANG: C++
 */
 #include<bits/stdc++.h>
 #define pb push_back
 #define FOR(i, n) for (int i = 0; i < (int)n; ++i)
 #define dbg(x) cout << #x << " at line " << __LINE__ << " is: " << x << endl
 typedef long long ll;
 using namespace std;
 typedef pair<int, int> pii;
 const int maxn = 1e5 + ;
 int n, m;
 int a[maxn * ];
 void solve() {
     int cur = ;
     cin >> n >> m;
     for (int i = ; i <= n; i++) {
         cin >> a[i];
         a[i + n] = a[i];
         if(i < m) cur += a[i];
     }
     int res = a[];
     for (int i = m; i <= n + m; i++) {
         cur += a[i];
         if(i - m > ) cur -= a[i - m];
         res = max(res, cur);
     }
     cout << res << endl;

 }
 int main() {
     //freopen("test.in", "r", stdin);
     //freopen("test.out", "w", stdout);
     int i; cin >> i;
     while(i--)
     solve();
     return ;
 }

2. 量子通讯工程，看完题目，就是写一个kruskal或者prim来求mst。 好久没写，发现都不会写了，刚好对dsu比较熟，就写了prim。

 /*
 ID: y1197771
 PROG: test
 LANG: C++
 */
 #include<bits/stdc++.h>
 #define pb push_back
 #define FOR(i, n) for (int i = 0; i < (int)n; ++i)
 #define dbg(x) cout << #x << " at line " << __LINE__ << " is: " << x << endl
 typedef long long ll;
 using namespace std;
 typedef pair<double, double> pii;
 const int maxn = 1e3 + ;
 double d[][];
 int n;
 vector<pair<double, double>> a;
 double work(pii x, pii y) {
     return sqrt((x.first - y.first) * (x.first - y.first) + (x.second - y.second) * ((x.second - y.second)) );
 }
 bool vis[];
 int f[maxn];
 int fd(int x) {
     if(x == f[x]) return f[x];
     return f[x] = fd(f[x]);
 }
 struct node {
     double d;
     int x, y;
     bool operator <(const node & t) const {
         return d < t.d;
     }
 } e[];
 void solve() {
     cin >> n;
     double x, y;
     for (int i = ; i < n; i++) {
         cin >> x >> y;
         a.pb({x, y});
         f[i] = i;
     }
     int num = ;
     for (int i = ; i < n; i++) {
         for (int j = i + ; j < n; j++) {
             double t = work(a[i], a[j]);
             e[num].d = t; e[num].x = i, e[num].y = j;
             num++;
         }
     }
     sort(e, e + num);
     double res = ;
     for (int i = ; i < num; i++) {
         int a1 = e[i].x, a2 = e[i].y;
         x = e[i].d;
         a1 = fd(a1); a2 = fd(a2);
         if(a1 != a2) {
             //cout << e[i].x << " " << e[i].y << " " << x << endl;
             f[a1] = a2;
             res += x;
         }
     }
     printf("%.2f\n", res);

 }
 int main() {
     //freopen("test.in", "r", stdin);
     //freopen("test.out", "w", stdout);
     solve();
     return ;
 }

也是直接过。

3. 下载管理软件，读题，主要是：任何下载的时候带宽都是满的，但是并行度个数有限制，然后我就想：直接模拟吧，按结束时间加入set进行模型，调了一个小时才过了33%。后来听师兄讲解，就是所有的任务加起来除以总带宽，那个并行度属于干扰项，我很惊讶，原来还可以这样！

写出来的有效代码不到5行，而我字节写的模拟，洋洋洒洒快到100行了。衰！

 /*
 ID: y1197771
 PROG: test
 LANG: C++
 */
 #include<bits/stdc++.h>
 #define pb push_back
 #define FOR(i, n) for (int i = 0; i < (int)n; ++i)
 #define dbg(x) cout << #x << " at line " << __LINE__ << " is: " << x << endl
 typedef long long ll;
 using namespace std;
 typedef pair<int, int> pii;
 const int maxn = 1e3 + ;
 void solve() {
     int t, n, w;
     double s = ;
     cin >> t >> n >> w;
     double task; int p;
     for (int i = ; i < t; i++) {
         cin >> task >> p;
         s += task * ( - p) / ;
     }
     printf("%.2f\n", s / w);
 }
 int main() {
     freopen("test.in", "r", stdin);
     //freopen("test.out", "w", stdout);
     solve();
     return ;
 }

http://ideone.com/ZqRTBn