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%。
1 /* 2 ID: y1197771 3 PROG: test 4 LANG: C++ 5 */ 6 #include<bits/stdc++.h> 7 #define pb push_back 8 #define FOR(i, n) for (int i = 0; i < (int)n; ++i) 9 #define dbg(x) cout << #x << " at line " << __LINE__ << " is: " << x << endl 10 typedef long long ll; 11 using namespace std; 12 typedef pair<int, int> pii; 13 const int maxn = 1e3 + 10; 14 string s[] = {"ONE", "TWO", "THREE", "FOUR", "FIVE", "SIX" }; 15 void solve() { 16 int n, a, b, c; 17 string s1, s2; 18 a = b = c = 0; 19 map<string, int> m; 20 cin >> n; 21 for (int i = 0; i < 6; i++) m[s[i] ] = i; 22 for (int i = 0; i < n; i++) { 23 cin >> s1 >> s2; 24 if(m[s1] > m[s2]) a++; 25 else if(m[s1] < m[s2]) b++; 26 else c++; 27 } 28 cout << a << " " << b << " " << c << endl; 29 if(b < a) { 30 cout << "MingMing Win" << endl; 31 } else { 32 cout << "LiangLiang Win" << endl; 33 } 34 35 } 36 int main() { 37 //freopen("test.in", "r", stdin); 38 //freopen("test.out", "w", stdout); 39 solve(); 40 return 0; 41 }
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。
1 /* 2 ID: y1197771 3 PROG: test 4 LANG: C++ 5 */ 6 #include<bits/stdc++.h> 7 #define pb push_back 8 #define FOR(i, n) for (int i = 0; i < (int)n; ++i) 9 #define dbg(x) cout << #x << " at line " << __LINE__ << " is: " << x << endl 10 typedef long long ll; 11 using namespace std; 12 typedef pair<int, int> pii; 13 const int maxn = 1e3 + 10; 14 int a[510][510]; 15 int cnt[510][510]; 16 struct node { 17 int v, x, y; 18 bool operator<(const node & t) const { 19 return v < t.v; 20 } 21 }; 22 int dx[] = {0, 1, 0, -1}; 23 int dy[] = {1, 0, -1, 0}; 24 void solve() { 25 set<node> se; 26 int n, m, x, y; cin >> n >> m; 27 for (int i = 0; i < n; i++) { 28 for (int j = 0; j < m; j++) { 29 cin >> x; a[i][j] = x; 30 se.insert({x, i, j}); 31 cnt[i][j] = 1; 32 } 33 } 34 int res = 0; 35 while(!se.empty()) { 36 int p = se.begin()->v; 37 x = se.begin()->x; y = se.begin()->y; 38 //cout << p << " " << x << " " << y << " " << res << endl; 39 res = max(res, cnt[x][y]); 40 se.erase(se.begin()); 41 for (int i = 0; i < 4; i++) { 42 int cx = x + dx[i], cy = y + dy[i]; 43 if(cx < 0 || cx >= n || cy < 0 || cy >= m) continue; 44 if(a[cx][cy] <= p) continue; 45 cnt[cx][cy] = max(cnt[x][y] + 1, cnt[cx][cy]); 46 } 47 } 48 cout << res << endl; 49 } 50 int main() { 51 freopen("test.in", "r", stdin); 52 //freopen("test.out", "w", stdout); 53 solve(); 54 return 0; 55 }
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。
1 /* 2 ID: y1197771 3 PROG: test 4 LANG: C++ 5 */ 6 #include<bits/stdc++.h> 7 #define pb push_back 8 #define FOR(i, n) for (int i = 0; i < (int)n; ++i) 9 #define dbg(x) cout << #x << " at line " << __LINE__ << " is: " << x << endl 10 typedef long long ll; 11 using namespace std; 12 typedef pair<int, int> pii; 13 const int maxn = 1e5 + 10; 14 int n, m; 15 int a[maxn * 2]; 16 void solve() { 17 int cur = 0; 18 cin >> n >> m; 19 for (int i = 1; i <= n; i++) { 20 cin >> a[i]; 21 a[i + n] = a[i]; 22 if(i < m) cur += a[i]; 23 } 24 int res = a[1]; 25 for (int i = m; i <= n + m; i++) { 26 cur += a[i]; 27 if(i - m > 0) cur -= a[i - m]; 28 res = max(res, cur); 29 } 30 cout << res << endl; 31 32 33 } 34 int main() { 35 //freopen("test.in", "r", stdin); 36 //freopen("test.out", "w", stdout); 37 int i; cin >> i; 38 while(i--) 39 solve(); 40 return 0; 41 }
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。
1 /* 2 ID: y1197771 3 PROG: test 4 LANG: C++ 5 */ 6 #include<bits/stdc++.h> 7 #define pb push_back 8 #define FOR(i, n) for (int i = 0; i < (int)n; ++i) 9 #define dbg(x) cout << #x << " at line " << __LINE__ << " is: " << x << endl 10 typedef long long ll; 11 using namespace std; 12 typedef pair<double, double> pii; 13 const int maxn = 1e3 + 10; 14 double d[110][110]; 15 int n; 16 vector<pair<double, double>> a; 17 double work(pii x, pii y) { 18 return sqrt((x.first - y.first) * (x.first - y.first) + (x.second - y.second) * ((x.second - y.second)) ); 19 } 20 bool vis[110]; 21 int f[maxn]; 22 int fd(int x) { 23 if(x == f[x]) return f[x]; 24 return f[x] = fd(f[x]); 25 } 26 struct node { 27 double d; 28 int x, y; 29 bool operator <(const node & t) const { 30 return d < t.d; 31 } 32 } e[10000]; 33 void solve() { 34 cin >> n; 35 double x, y; 36 for (int i = 0; i < n; i++) { 37 cin >> x >> y; 38 a.pb({x, y}); 39 f[i] = i; 40 } 41 int num = 0; 42 for (int i = 0; i < n; i++) { 43 for (int j = i + 1; j < n; j++) { 44 double t = work(a[i], a[j]); 45 e[num].d = t; e[num].x = i, e[num].y = j; 46 num++; 47 } 48 } 49 sort(e, e + num); 50 double res = 0; 51 for (int i = 0; i < num; i++) { 52 int a1 = e[i].x, a2 = e[i].y; 53 x = e[i].d; 54 a1 = fd(a1); a2 = fd(a2); 55 if(a1 != a2) { 56 //cout << e[i].x << " " << e[i].y << " " << x << endl; 57 f[a1] = a2; 58 res += x; 59 } 60 } 61 printf("%.2f ", res); 62 63 } 64 int main() { 65 //freopen("test.in", "r", stdin); 66 //freopen("test.out", "w", stdout); 67 solve(); 68 return 0; 69 }
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！
1 /* 2 ID: y1197771 3 PROG: test 4 LANG: C++ 5 */ 6 #include<bits/stdc++.h> 7 #define pb push_back 8 #define FOR(i, n) for (int i = 0; i < (int)n; ++i) 9 #define dbg(x) cout << #x << " at line " << __LINE__ << " is: " << x << endl 10 typedef long long ll; 11 using namespace std; 12 typedef pair<int, int> pii; 13 const int maxn = 1e3 + 10; 14 void solve() { 15 int t, n, w; 16 double s = 0; 17 cin >> t >> n >> w; 18 double task; int p; 19 for (int i = 0; i < t; i++) { 20 cin >> task >> p; 21 s += task * (100 - p) / 100; 22 } 23 printf("%.2f ", s / w); 24 } 25 int main() { 26 freopen("test.in", "r", stdin); 27 //freopen("test.out", "w", stdout); 28 solve(); 29 return 0; 30 }