没想到平成年代最后一场cf居然是手速场，幸好拿了个小号娱乐不然掉分预定 (逃
题目链接：http://codeforces.com/contest/1150
A：
傻题。
1 /* basic header */ 2 #include <iostream> 3 #include <cstdio> 4 #include <cstdlib> 5 #include <string> 6 #include <cstring> 7 #include <cmath> 8 #include <cstdint> 9 #include <climits> 10 #include <float.h> 11 /* STL */ 12 #include <vector> 13 #include <set> 14 #include <map> 15 #include <queue> 16 #include <stack> 17 #include <algorithm> 18 #include <array> 19 #include <iterator> 20 /* define */ 21 #define ll long long 22 #define dou double 23 #define pb emplace_back 24 #define mp make_pair 25 #define fir first 26 #define sec second 27 #define sot(a,b) sort(a+1,a+1+b) 28 #define rep1(i,a,b) for(int i=a;i<=b;++i) 29 #define rep0(i,a,b) for(int i=a;i<b;++i) 30 #define repa(i,a) for(auto &i:a) 31 #define eps 1e-8 32 #define int_inf 0x3f3f3f3f 33 #define ll_inf 0x7f7f7f7f7f7f7f7f 34 #define lson curPos<<1 35 #define rson curPos<<1|1 36 /* namespace */ 37 using namespace std; 38 /* header end */ 39 40 const int maxn = 1e2 + 10; 41 int n, m, r, minn = 2000, maxx = 0; 42 43 int main() 44 { 45 scanf("%d%d%d", &n, &m, &r); 46 rep1(i, 1, n) 47 { 48 int x; scanf("%d", &x); minn = min(minn, x); 49 } 50 rep1(i, 1, m) 51 { 52 int x; scanf("%d", &x); maxx = max(maxx, x); 53 } 54 if (maxx > minn) printf("%d ", maxx * (r / minn) + r % minn); 55 else printf("%d ", r); 56 return 0; 57 }
B：
乍一看以为要搜，其实根本不需要，n^2填进去判断就完事了 (看通过人数就能猜到根本不用搜。
至于为什么可以这样，是因为跟紧密填充相关吗？
1 /* basic header */ 2 #include <iostream> 3 #include <cstdio> 4 #include <cstdlib> 5 #include <string> 6 #include <cstring> 7 #include <cmath> 8 #include <cstdint> 9 #include <climits> 10 #include <float.h> 11 /* STL */ 12 #include <vector> 13 #include <set> 14 #include <map> 15 #include <queue> 16 #include <stack> 17 #include <algorithm> 18 #include <array> 19 #include <iterator> 20 /* define */ 21 #define ll long long 22 #define dou double 23 #define pb emplace_back 24 #define mp make_pair 25 #define fir first 26 #define sec second 27 #define sot(a,b) sort(a+1,a+1+b) 28 #define rep1(i,a,b) for(int i=a;i<=b;++i) 29 #define rep0(i,a,b) for(int i=a;i<b;++i) 30 #define repa(i,a) for(auto &i:a) 31 #define eps 1e-8 32 #define int_inf 0x3f3f3f3f 33 #define ll_inf 0x7f7f7f7f7f7f7f7f 34 #define lson curPos<<1 35 #define rson curPos<<1|1 36 /* namespace */ 37 using namespace std; 38 /* header end */ 39 40 const int maxn = 55; 41 int a[maxn][maxn], n, solved = 0; 42 char s[maxn]; 43 44 int check() 45 { 46 rep1(i, 1, n) 47 rep1(j, 1, n) 48 if (!a[i][j]) return 0; 49 return 1; 50 } 51 52 int putable(int x, int y) 53 { 54 if (x == 1 || x == n || y == 1 || y == n) return 0; 55 if (a[x - 1][y] || a[x + 1][y] || a[x][y - 1] || a[x][y + 1] || a[x][y]) return 0; 56 return 1; 57 } 58 59 int main() 60 { 61 scanf("%d", &n); 62 rep1(i, 1, n) 63 { 64 scanf("%s", s + 1); 65 rep1(j, 1, n) 66 if (s[j] == '#') a[i][j] = 1; else a[i][j] = 0; 67 } 68 rep1(i, 2, n - 1) 69 { 70 rep1(j, 2, n - 1) 71 if (putable(i, j)) 72 { 73 a[i - 1][j] = a[i + 1][j] = a[i][j - 1] = a[i][j + 1] = a[i][j] = 1; 74 } 75 } 76 if (check()) puts("YES"); 77 else puts("NO"); 78 return 0; 79 }
C：
2e5质数筛贪心就完事了，优先满足最近的质数，先塞2再塞1，也有大佬说不用筛。
1 /* basic header */ 2 #include <iostream> 3 #include <cstdio> 4 #include <cstdlib> 5 #include <string> 6 #include <cstring> 7 #include <cmath> 8 #include <cstdint> 9 #include <climits> 10 #include <float.h> 11 /* STL */ 12 #include <vector> 13 #include <set> 14 #include <map> 15 #include <queue> 16 #include <stack> 17 #include <algorithm> 18 #include <array> 19 #include <iterator> 20 /* define */ 21 #define ll long long 22 #define dou double 23 #define pb emplace_back 24 #define mp make_pair 25 #define fir first 26 #define sec second 27 #define sot(a,b) sort(a+1,a+1+b) 28 #define rep1(i,a,b) for(int i=a;i<=b;++i) 29 #define rep0(i,a,b) for(int i=a;i<b;++i) 30 #define repa(i,a) for(auto &i:a) 31 #define eps 1e-8 32 #define int_inf 0x3f3f3f3f 33 #define ll_inf 0x7f7f7f7f7f7f7f7f 34 #define lson curPos<<1 35 #define rson curPos<<1|1 36 /* namespace */ 37 using namespace std; 38 /* header end */ 39 40 const int maxn = 2e5 + 10; 41 int c1 = 0, c2 = 0, n, prime[maxn], p = 1, tot; 42 vector<int>ans; 43 44 bool valid[maxn]; 45 void getPrime(int n, int &tot, int ans[maxn]) 46 { 47 tot = 0; 48 memset(valid, 1, sizeof(valid)); 49 for (int i = 2; i <= n; i++) 50 { 51 if (valid[i]) ans[++tot] = i; 52 for (int j = 1; (j <= tot) && (i * ans[j] <= n); j++) 53 { 54 valid[i * ans[j]] = false; 55 if (i % ans[j] == 0) break; 56 } 57 } 58 } 59 60 int main() 61 { 62 getPrime(2e5, tot, prime); 63 ans.clear(); 64 scanf("%d", &n); 65 rep1(i, 1, n) 66 { 67 int x; scanf("%d", &x); 68 if (x & 1) c1++; else c2++; 69 } 70 int curr = 0; 71 while (c1 || c2) 72 { 73 if (prime[p] - curr > 2 && c2) 74 { 75 ans.pb(2); c2--; curr += 2; 76 } 77 else if (prime[p] - curr == 2 && c2) 78 { 79 ans.pb(2); c2--; p++; curr += 2; 80 } 81 else if (prime[p] - curr == 1 && c1) 82 { 83 ans.pb(1); c1--; p++; curr += 1; 84 } 85 else 86 { 87 if (c2) 88 { 89 ans.pb(2); c2--; curr += 2; 90 } 91 else if (c1) 92 { 93 ans.pb(1); c1--; curr += 1; 94 } 95 } 96 } 97 for (auto i : ans) printf("%d ", i); 98 return 0; 99 }
D：
把全场人卡死的dp，直到比赛结束都只有不到20个人过了.
设f[i][j][k]表示第1个串匹配到i，第2个串匹配到j，第3个串匹配到k所用的最小长度，然后每次添加一个字符可以O(250^2)转移另外两维的状态即可。
复杂度：O(q*250^2)
E：
给定一个括号序列，每个括号序列可以唯一生成一棵有根树。给出q次询问，每次询问交换上次得到的括号序列中某两个括号的位置（保证交换之后仍然可以生成一棵有根树）。输出交换之后得到的有根树的直径。
一看就是线段树可以做的题。
1 /* basic header */ 2 #include <iostream> 3 #include <cstdio> 4 #include <cstdlib> 5 #include <string> 6 #include <cstring> 7 #include <cmath> 8 #include <cstdint> 9 #include <climits> 10 #include <float.h> 11 /* STL */ 12 #include <vector> 13 #include <set> 14 #include <map> 15 #include <queue> 16 #include <stack> 17 #include <algorithm> 18 #include <array> 19 #include <iterator> 20 /* define */ 21 #define ll long long 22 #define dou double 23 #define pb emplace_back 24 #define mp make_pair 25 #define fir first 26 #define sec second 27 #define sot(a,b) sort(a+1,a+1+b) 28 #define rep1(i,a,b) for(int i=a;i<=b;++i) 29 #define rep0(i,a,b) for(int i=a;i<b;++i) 30 #define repa(i,a) for(auto &i:a) 31 #define eps 1e-8 32 #define int_inf 0x3f3f3f3f 33 #define ll_inf 0x7f7f7f7f7f7f7f7f 34 #define lson curPos<<1 35 #define rson curPos<<1|1 36 /* namespace */ 37 using namespace std; 38 /* header end */ 39 40 const int maxn = 2e5 + 10; 41 char s[maxn]; 42 43 struct Node 44 { 45 int sum, pmx, lmx, minn, maxx, d; 46 Node() {} 47 Node(int a, int b, int c, int d, int e, int f) 48 { 49 sum = a; pmx = b; lmx = c; minn = d; maxx = e; d = f; 50 } 51 }; 52 53 //ostream &operator<<(ostream &os, const Node &rhs) 54 //{ 55 // os << rhs.sum << " " << rhs.pmx << " " << rhs.lmx << " " << rhs.minn << " " << rhs.maxx << " " << rhs.d; 56 // return os; 57 //} 58 59 Node segT[maxn << 2]; 60 61 void maintain(int curPos) 62 { 63 segT[curPos].sum = segT[lson].sum + segT[rson].sum; 64 segT[curPos].pmx = max(segT[lson].pmx, max(segT[rson].maxx - 2 * segT[lson].minn + segT[lson].sum, segT[rson].pmx - segT[lson].sum)); 65 segT[curPos].lmx = max(segT[lson].lmx, max(segT[rson].lmx - segT[lson].sum, segT[lson].maxx - 2 * segT[rson].minn - 2 * segT[lson].sum)); 66 segT[curPos].minn = min(segT[lson].minn, segT[rson].minn + segT[lson].sum); 67 segT[curPos].maxx = max(segT[lson].maxx, segT[rson].maxx + segT[lson].sum); 68 segT[curPos].d = max(max(segT[lson].d, segT[rson].d), max(segT[lson].maxx + segT[rson].pmx - segT[lson].sum, segT[lson].lmx + segT[lson].sum + segT[rson].maxx)); 69 } 70 71 void build(int curPos, int curL, int curR) 72 { 73 if (curL == curR) 74 { 75 int x = s[curL] == '(' ? 1 : -1; 76 segT[curPos] = Node(x, -x, -x, x, x, 0); 77 // cout << segT[curPos] << endl; 78 return; 79 } 80 int mid = curL + curR >> 1; 81 build(lson, curL, mid); build(rson, mid + 1, curR); 82 maintain(curPos); 83 } 84 85 void update(int pos, int curPos, int curL, int curR) 86 { 87 if (curL == curR) 88 return build(curPos, curL, curR); 89 int mid = curL + curR >> 1; 90 if (pos <= mid) 91 update(pos, lson, curL, mid); 92 else 93 update(pos, rson, mid + 1, curR); 94 maintain(curPos); 95 } 96 97 int main() 98 { 99 int n, m; scanf("%d%d", &n, &m); 100 scanf("%s", s + 1); 101 n = n * 2 - 2; 102 build(1, 1, n); 103 printf("%d ", segT[1].d); 104 while (m--) 105 { 106 int x, y; scanf("%d%d", &x, &y); 107 swap(s[x], s[y]); 108 update(x, 1, 1, n); update(y, 1, 1, n); 109 printf("%d ", segT[1].d); 110 } 111 return 0; 112 }