今天我挺有状态的，看过的题基本都给了正解(可能是昨晚cf div3打得跟屎一样，人品守恒，不好意思发题解了)，自己也给队伍签了很多水题(不敢让队友写，怕出锅)。
最后6题滚了，有点可惜。还差B和K没做出来。
B我一眼就知道该怎么做了，推了一会发现相当可做，于是给队友嘴巴完队友写挂了 (有些小细节，比如判0什么的，下面的题解会讲)。
K有不同区间的做法一眼秒，没有的话就manacher，队友又写挂了 (弟弟们，怎么回事？
今天的训练我留意到一件事情，我跟队友讲题意的时候他们能听懂，但他们跟我讲题意的时候往往要讲好几遍。只要描述的过程中出现类似“这个”“那个”这样的代词我就完全不能接受，一定会让队友明确指出是什么东西。
以后训练我想刻意地让队友来讲题意，让他们多锻炼一下表达能力。
题目链接：http://acm.zju.edu.cn/onlinejudge/showContestProblems.do?contestId=392
A：
图论题。DSU+权值线段树+二分查询+读优才能过。考验码力。
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 = 1e5 + 10; 41 ll sum[maxn << 2], sumn[maxn << 2], sume[maxn << 2], nei[maxn << 2], summan[maxn << 2]; 42 int f[maxn], fsz[maxn], n, Q; 43 ll fed[maxn], sn, sm, tn; 44 45 inline ll read() 46 { 47 ll d = 0; 48 char s = getchar(); 49 while (s < '0' || s > '9') 50 s = getchar(); 51 while (s >= '0' && s <= '9') 52 { 53 d = d * 10 + s - '0'; 54 s = getchar(); 55 } 56 return d; 57 } 58 59 int findFa(int x) 60 { 61 if (f[x] == x) return x; 62 return f[x] = findFa(f[x]); 63 } 64 65 void maintain(int curPos) 66 { 67 sum[curPos] = sum[lson] + sum[rson]; 68 sumn[curPos] = sumn[lson] + sumn[rson]; 69 sume[curPos] = sume[lson] + sume[rson]; 70 nei[curPos] = nei[lson] + nei[rson]; 71 summan[curPos] = summan[lson] + summan[rson]; 72 } 73 74 void build(int curPos, int curL, int curR) 75 { 76 if (curL == curR) 77 { 78 sumn[curPos] = sum[curPos] = curL == 1 ? n : 0; 79 sume[curPos] = nei[curPos] = summan[curPos] = 0; 80 return; 81 } 82 int mid = (curL + curR) >> 1; 83 build(lson, curL, mid); build(rson, mid + 1, curR); 84 maintain(curPos); 85 } 86 87 void add(int pos, int val, int curPos, int curL, int curR, ll ve) 88 { 89 if (curL == curR) 90 { 91 sum[curPos] += (ll)curL * val; 92 sumn[curPos] += val; 93 sume[curPos] += ve; 94 nei[curPos] = (ll)curL * (curL - 1) / 2 * sumn[curPos] - sume[curPos]; 95 summan[curPos] = (ll)curL * (curL - 1) / 2 * sumn[curPos]; 96 return; 97 } 98 int mid = (curL + curR) >> 1; 99 if (pos <= mid) 100 add(pos, val, lson, curL, mid, ve); 101 else 102 add(pos, val, rson, mid + 1, curR, ve); 103 maintain(curPos); 104 } 105 106 void query(int curPos, int l, int r, ll K) 107 { 108 if (l == r) 109 { 110 int _l = 1, _r = sumn[curPos], ans = K ? 1 : 0; 111 while (_l <= _r) 112 { 113 int mid = (_l + _r) >> 1; 114 ll sp = sn + (ll) l * mid; 115 ll g = sp * (sp - 1) / 2 - sm - (ll) l * (l - 1) / 2 * mid; 116 // cerr << mid << ' ' << g << ' '; 117 if (g <= K) 118 { 119 if (g < K) 120 ans = max(ans, mid + 1); 121 else 122 ans = max(ans, mid); 123 _l = mid + 1; 124 } 125 else 126 _r = mid - 1; 127 } 128 tn += ans; 129 return; 130 } 131 int mid = (l + r) >> 1; 132 ll sp = sn + sum[rson]; 133 ll g = sp * (sp - 1) / 2 - summan[rson] - sm; 134 if (g == K) 135 { 136 sn += sum[rson]; 137 tn += sumn[rson]; 138 } 139 if (g < K) 140 { 141 sn += sum[rson]; 142 sm += summan[rson]; 143 tn += sumn[rson]; 144 query(lson, l, mid, K); 145 } 146 if (g > K) 147 query(rson, mid + 1, r, K); 148 } 149 150 int main() 151 { 152 int t=read(); 153 while (t--) 154 { 155 n=read(),Q=read(); 156 rep1(i, 1, n) 157 { 158 f[i] = i, fsz[i] = 1, fed[i] = 0; 159 } 160 build(1, 1, n); 161 int bsz = n; 162 while (Q--) 163 { 164 int op=read(); 165 if (op == 1) 166 { 167 int x=read(), y=read(); 168 int dx = findFa(x), dy = findFa(y); 169 if (dx == dy) 170 { 171 fed[dx]++; 172 add(fsz[dx], 0, 1, 1, n, 1); 173 } 174 else 175 { 176 add(fsz[dx], -1, 1, 1, n, -fed[dx]); 177 add(fsz[dy], -1, 1, 1, n, -fed[dy]); 178 add(fsz[dx] + fsz[dy], 1, 1, 1, n, fed[dx] + fed[dy] + 1); 179 f[dx] = dy; 180 fsz[dy] += fsz[dx]; 181 fed[dy] += fed[dx] + 1; 182 bsz--; 183 } 184 } 185 else 186 { 187 ll K=read(); 188 int mi = max(bsz - K, 1LL), ma; 189 K -= nei[1]; 190 if (K <= 0) 191 ma = bsz; 192 else 193 { 194 sn = sm = tn = 0; 195 query(1, 1, n, K); 196 ma = bsz - tn + 1; 197 } 198 printf("%d %d ", mi, ma); 199 } 200 } 201 } 202 return 0; 203 }
不上读优就是这样的下场（1920ms那次A得莫名其妙）：
B：
定义x,y为原始数组(样例已经给定)的值，x',y'为给定数组计算出来的值。显然可以作x-x',y-y'并提取公因式，计算一下(y-y')/(x-x')==a[i]+a[j]。到这里，解法已经很清晰了：枚举i，求j和a[j]并验证之。
细节有点恶心，要判x-x',y-y'是否为0。仔细点就可以过了。
/* basic header */ #include <iostream> #include <cstdio> #include <cstdlib> #include <string> #include <cstring> #include <cmath> #include <cstdint> #include <climits> #include <float.h> /* STL */ #include <vector> #include <set> #include <map> #include <queue> #include <stack> #include <algorithm> #include <array> #include <iterator> /* define */ #define ll long long #define dou double #define pb emplace_back #define mp make_pair #define fir first #define sec second #define init(a,b) fill(begin(a),end(a),b) #define sot(a,b) sort(a+1,a+1+b) #define rep1(i,a,b) for(int i=a;i<=b;++i) #define rep0(i,a,b) for(int i=a;i<b;++i) #define repa(i,a) for(auto &i:a) #define eps 1e-8 #define int_inf 0x3f3f3f3f #define ll_inf 0x7f7f7f7f7f7f7f7f #define lson curPos<<1 #define rson curPos<<1|1 /* namespace */ using namespace std; /* header end */ const int maxn = 1e5 + 10; ll a[maxn], b[maxn], cnt[maxn]; int t; int main() { scanf("%d", &t); while (t--) { ll n, x, y, curX = 0, curY = 0; scanf("%lld%lld%lld", &n, &x, &y); rep1(i, 1, n) { scanf("%lld", &a[i]); b[i - 1] = a[i]; cnt[i - 1] = 0; curX += (i * a[i]); curY += (i * a[i] * a[i]); } ll deltaX = x - curX, deltaY = y - curY; if (!deltaX && deltaY) //除0说明无解 { puts("0"); continue; } if (!deltaX && !deltaY) //这情况就很有意思了 { ll ans = 0; sort(b, b + n); int m = unique(b, b + n) - b; rep1(i, 1, n) cnt[lower_bound(b, b + m, a[i]) - b]++; rep0(i, 0, m) ans += 1LL * cnt[i] * (cnt[i] - 1) / 2; printf("%lld ", ans); continue; } //不能整除说明无解，不能跟上面的if合在一起，不然会报浮点错误 if (abs(deltaY) % abs(deltaX)) { puts("0"); continue; } //有解，枚举i即可 ll rough = deltaY / deltaX, ans = 0; for (ll i = 1; i <= n; i++) { ll fm = rough - 2 * a[i], fs = deltaX / fm, j = i - fs; if (!fm) continue; if (abs(deltaX) % abs(fm)) continue; if (j <= n && j >= 1) if (a[i] + a[j] == rough) ans++; } printf("%lld ", ans / 2); } return 0; }
C：
这个题给队友喂了好多假算法，都被队友hack掉了，有点迷。当时感觉是个姿势非常奇怪的贪心。
晚上写完这个博客之后洗了个澡，突然明白这题是从前往后贪心，线段树维护。(当时看到n<=1e5我就觉得是Onlogn的算法)
然而过了两天之后我发现了更好的做法，发现这个题只要维护得好，On就可以做了。
不得不承认确实是个出得不错的贪心，看完代码甚至觉得这题非常傻……
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 = 1e5 + 10; 41 int n, a[maxn], cnt[maxn], cntTwi[maxn], ans[maxn]; //cntTwi[i]==cnt[i]*2 42 set<pair<int, int>>s; //次数的两倍和数值makepair，set的特性会先按cnt排序，再按数值排序 43 set<int>all; 44 45 void init(int n) 46 { 47 s.clear(); all.clear(); 48 rep1(i, 1, n) cnt[i] = cntTwi[i] = ans[i] = 0; 49 } 50 51 int main() 52 { 53 int t; scanf("%d", &t); 54 while (t--) 55 { 56 scanf("%d", &n); 57 init(n); 58 for (int i = 1; i <= n; i++) 59 { 60 scanf("%d", &a[i]); 61 cnt[a[i]]++; 62 } 63 for (int i = 1; i <= n; i++) 64 if (cnt[i]) 65 { 66 all.insert(i); 67 s.emplace(cntTwi[i] = 2 * cnt[i], i); 68 } 69 int flag = 0; //无解标志 70 for (int i = 1; !flag && i <= n; i++) 71 { 72 int p = s.rbegin()->second; //先处理cnt最多而且数值较大的那个数字 73 if (cntTwi[p] > n - i + 1) //如果cnt大于剩余位置的一半，必然无解 74 { 75 flag = 1; 76 puts("Impossible"); 77 break; 78 } 79 else if (cntTwi[p] < n - i + 1 || p == a[i]) //如果要处理的数字正好for到了，不能放下去，那就找个最小的可用的数字替代之 80 { 81 auto it = all.begin(); 82 if (*it == a[i]) it++; 83 p = *it; 84 } 85 //更新re 86 s.erase({cntTwi[a[i]], a[i]}); 87 s.erase({cntTwi[p], p}); 88 if (--cntTwi[a[i]]) s.emplace(cntTwi[a[i]], a[i]); 89 if (--cntTwi[p]) s.emplace(cntTwi[p], p); 90 if (--cnt[ans[i] = p] == 0) all.erase(p); 91 } 92 if (!flag) 93 for (int i = 1; i <= n; i++) 94 printf("%d%c", ans[i], i < n ? ' ' : ' '); 95 } 96 return 0; 97 }
D：
有向图求欧拉路（貌似？我还挺想写的，但是最后的时间都在调B和K，难受。
补题的时候发现这题并没有我想得这么简单，幸好比赛的时候没写，不然更浪费时间了。
原本我还想着什么记忆化之类的事情，dalao直接一发暴力就A了。
Input最后有一句“It's guaranteed that the sum of of all test cases will not exceed 10^6”。当时没看到……
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 struct Graph 41 { 42 struct Vertex 43 { 44 vector<int>to; 45 int inDeg, vis; //inDeg:入度 46 }; 47 vector<Vertex>vertex; 48 vector<int>eulerPath; 49 Graph(int n): vertex(n) {} 50 51 int dfs(int u, int dep) 52 { 53 vertex[u].vis = 1; 54 eulerPath.pb(u); 55 if (!dep) return 1; 56 for (auto to : vertex[u].to) 57 if (!vertex[to].vis) 58 { 59 int flag = 1; 60 vertex[to].inDeg++; 61 for (auto to : vertex[u].to) 62 if (!--vertex[to].inDeg && !vertex[to].vis) flag = 0; 63 if (flag && dfs(to, dep - 1)) 64 return 1; 65 vertex[to].inDeg--; 66 for (auto to : vertex[u].to) vertex[to].inDeg++; 67 } 68 //如果for一遍都没结果，说明不存在欧拉路 69 eulerPath.pop_back(); 70 vertex[u].vis = 0; 71 return 0; 72 } 73 }; 74 75 int main() 76 { 77 int t; scanf("%d", &t); 78 while (t--) 79 { 80 int n; scanf("%d", &n); 81 Graph g(n + 1); 82 for (int i = 1; i <= n; i++) //建图 83 for (auto to : 84 { 85 i / 2, i - 1, i * 2, i * 2 + 1 86 }) 87 if (2 <= to && to <= n) 88 g.vertex[i].to.pb(to), g.vertex[to].inDeg++; 89 if (!g.dfs(1, n - 1)) 90 puts("Impossible"); 91 else 92 for (int i = 0; i < g.eulerPath.size(); i++) 93 printf("%d%c", g.eulerPath[i], i < g.eulerPath.size() - 1 ? ' ' : ' '); 94 } 95 return 0; 96 }
E：
sort一下，倒着判就完事了。
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 init(a,b) fill(begin(a),end(a),b) 28 #define sot(a,b) sort(a+1,a+1+b) 29 #define rep1(i,a,b) for(int i=a;i<=b;++i) 30 #define rep0(i,a,b) for(int i=a;i<b;++i) 31 #define repa(i,a) for(auto &i:a) 32 #define eps 1e-8 33 #define int_inf 0x3f3f3f3f 34 #define ll_inf 0x7f7f7f7f7f7f7f7f 35 #define lson curPos<<1 36 #define rson curPos<<1|1 37 /* namespace */ 38 using namespace std; 39 /* header end */ 40 41 const int maxn = 1e5 + 10; 42 int n, a[maxn], b[maxn], ans; 43 44 int main() 45 { 46 int t; scanf("%d", &t); 47 while (t--) 48 { 49 scanf("%d", &n); 50 rep1(i, 1, n) scanf("%d", &a[i]), b[i] = a[i]; 51 sot(b, n); 52 ans = n; 53 for (int i = n; i >= 1; i--) 54 if (a[i] == b[ans]) ans--; 55 printf("%d ", ans); 56 } 57 return 0; 58 }
F：
没什么意思的细节签到题。
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 init(a,b) fill(begin(a),end(a),b) 28 #define sot(a,b) sort(a+1,a+1+b) 29 #define rep1(i,a,b) for(int i=a;i<=b;++i) 30 #define rep0(i,a,b) for(int i=a;i<b;++i) 31 #define repa(i,a) for(auto &i:a) 32 #define eps 1e-8 33 #define int_inf 0x3f3f3f3f 34 #define ll_inf 0x7f7f7f7f7f7f7f7f 35 #define lson curPos<<1 36 #define rson curPos<<1|1 37 /* namespace */ 38 using namespace std; 39 /* header end */ 40 41 set<char>ss = { 'a', 'e', 'i', 'y', 'o', 'u' }; 42 const int maxn = 200; 43 char s[maxn]; 44 int t; 45 46 int main() 47 { 48 scanf("%d", &t); 49 while (t--) 50 { 51 scanf("%s", s + 1); 52 int len = strlen(s + 1); 53 printf("%c", s[1]); 54 rep1(i, 2, len) 55 if (ss.count(s[i])) continue; 56 else printf("%c", s[i]); 57 puts(""); 58 } 59 return 0; 60 }
G：
温暖人心。
1 #include <bits/stdc++.h> 2 /* define */ 3 #define ll long long 4 #define dou double 5 #define pb emplace_back 6 #define mp make_pair 7 #define fir first 8 #define sec second 9 #define init(a,b) fill(begin(a),end(a),b) 10 #define sot(a,b) sort(a+1,a+1+b) 11 #define rep1(i,a,b) for(int i=a;i<=b;++i) 12 #define rep0(i,a,b) for(int i=a;i<b;++i) 13 #define repa(i,a) for(auto &i:a) 14 #define eps 1e-8 15 #define int_inf 0x3f3f3f3f 16 #define ll_inf 0x7f7f7f7f7f7f7f7f 17 #define lson curPos<<1 18 #define rson curPos<<1|1 19 /* namespace */ 20 using namespace std; 21 /* header end */ 22 23 int t, n; 24 25 int main() 26 { 27 scanf("%d", &t); 28 while (t--) 29 { 30 scanf("%d", &n); 31 for (int i = n;; i++) 32 { 33 if (i % 7 == 0 && i % 4 != 0) 34 { 35 printf("%d ", i); 36 break; 37 } 38 } 39 } 40 return 0; 41 }
H：
水题，计算出原有极值个数p，答案显然只有p-0,p-1,p-2三种情况（样例太恶臭了。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 5 using namespace std; 6 const int maxn = 2000000; 7 int A[maxn]; 8 int fg[maxn]; 9 int main() 10 { 11 int T; 12 scanf("%d",&T); 13 while(T--){ 14 int n; 15 scanf("%d",&n); 16 for(int i = 1;i <= n;i++){ 17 scanf("%d",&A[i]); 18 fg[i] = 0; 19 } 20 if(n <= 3){ 21 printf("0 "); 22 continue; 23 } 24 int ans = 0; 25 for(int i = 2;i <= n-1;i++){ 26 if(A[i-1] < A[i] &&A[i] > A[i+1]){ 27 ans ++; 28 fg[i] = 1; 29 } 30 } 31 // for(int i = 1;i <= n;i++) 32 // printf("%d ",A[i]); 33 // printf(" "); 34 // for(int i = 1;i <= n;i++) 35 // printf("%d ",fg[i]); 36 // printf(" "); 37 int endflag = 0; 38 for(int i = 2;i <= n-2;i++){ 39 if(fg[i] == 1&&fg[i + 2] == 1){ 40 if(A[i] == A[i + 2]){ 41 printf("%d ",ans-2); 42 endflag = 1; 43 break; 44 } 45 } 46 } 47 if(endflag) continue; 48 for(int i = 2;i <= n-2;i++){ 49 if(fg[i] == 1){ 50 if(A[i-1] < A[i+1]&&A[i+1] > A[i+2]){ 51 continue; 52 }else if(A[i-2] < A[i-1]&&A[i-1] > A[i+1]) { 53 continue; 54 }else{ 55 printf("%d ",ans-1); 56 endflag = 1; 57 break; 58 } 59 } 60 } 61 if(endflag) continue; 62 printf("%d ",ans); 63 } 64 return 0; 65 }
I：
看完题之后队友马上给结论：判读进来的数模3余几即可，虽然我没有听得很懂，不过看队友一脸坚定我还是光速敲了。奇偶性貌似是看能不能被3整除，研究一下就知道了。
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 init(a,b) fill(begin(a),end(a),b) 28 #define sot(a,b) sort(a+1,a+1+b) 29 #define rep1(i,a,b) for(int i=a;i<=b;++i) 30 #define rep0(i,a,b) for(int i=a;i<b;++i) 31 #define repa(i,a) for(auto &i:a) 32 #define eps 1e-8 33 #define int_inf 0x3f3f3f3f 34 #define ll_inf 0x7f7f7f7f7f7f7f7f 35 #define lson curPos<<1 36 #define rson curPos<<1|1 37 /* namespace */ 38 using namespace std; 39 /* header end */ 40 41 const int maxn = 1e4 + 10; 42 char a[maxn], b[maxn]; 43 44 int main() 45 { 46 int t; 47 scanf("%d", &t); 48 while (t--) 49 { 50 scanf("%s", a + 1); scanf("%s", b + 1); 51 int len1 = strlen(a + 1), len2 = strlen(b + 1); 52 int sum1 = 0, sum2 = 0, sign1 = 1, sign2 = 1; 53 rep1(i, 1, len1) sum1 += a[i] - '0'; 54 rep1(i, 1, len2) sum2 += b[i] - '0'; 55 if (sum1 % 3 == 2) sign1 = 2; if (sum2 % 3 == 1) sign2 = 2; 56 printf("%d ", abs(sign1 - sign2) % 2); 57 } 58 return 0; 59 }
J：
DSU水题，队友居然写挂了要批评，还是我改对的。
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int maxn = 2000000; 4 int pa[maxn]; 5 6 int find(int x) 7 { 8 if (pa[x] == x) return x; 9 else return pa[x] = find(pa[x]); 10 } 11 priority_queue<int, vector<int>, greater<int> > Q; 12 vector<int> G[maxn]; 13 int vis[maxn], visit[maxn]; 14 15 int main() 16 { 17 int T; 18 scanf("%d", &T); 19 while (T--) 20 { 21 int n, m; 22 scanf("%d%d", &n, &m); 23 for (int i = 1; i <= n; i++) 24 { 25 pa[i] = i; 26 vis[i] = 0; 27 visit[i] = 0; 28 G[i].clear(); 29 } 30 for (int i = 1; i <= m; i++) 31 { 32 int a, b; 33 scanf("%d%d", &a, &b); 34 if (a > b) swap(a, b); 35 G[a].push_back(b); 36 G[b].push_back(a); 37 int x = find(a); 38 int y = find(b); 39 pa[y] = x; 40 } 41 int ans = 0; 42 for (int i = 1; i <= n; i++) 43 { 44 if (find(i) == i) ans ++; 45 } 46 printf("%d ", ans); 47 while (!Q.empty()) Q.pop(); 48 for (int i = 1; i <= n; i++) 49 { 50 if (vis[pa[i]] == 0) 51 { 52 Q.push(i); 53 vis[pa[i]] = 1; 54 visit[i] = 1; 55 } 56 } 57 int flag = 0; 58 while (!Q.empty()) 59 { 60 int x = Q.top(); 61 Q.pop(); 62 if (flag == 0) 63 { 64 printf("%d", x); 65 flag = 1; 66 } 67 else printf(" %d", x); 68 for (int i = 0; i < G[x].size(); i++) 69 { 70 if (!visit[G[x][i]]) 71 { 72 visit[G[x][i]] = 1; 73 Q.push(G[x][i]); 74 } 75 } 76 } 77 printf(" "); 78 } 79 return 0; 80 }
K：
和队友讨论了一会口A了。如果存在不相同子串的区间，那么选定一个尽量小的“能包含所有不相同子串区间”的区间，然后往两边扩展即可。
如果不存在不相同子串的区间，直接跑manacher算回文子串个数就完事了。 (赛后队友原来是死在ans没开LL，队友锅++
1 #include <cstdio> 2 #include <string> 3 #include <iostream> 4 #include <algorithm> 5 6 using namespace std; 7 8 const int maxn = 2e6+1; 9 10 char ma[maxn * 2]; 11 int mp[maxn * 2]; 12 13 long long int manacher (string s) { 14 auto len = s.length(); 15 int l = 0; 16 ma[l++] = '$'; 17 ma[l++] = '#'; 18 for (int i = 0; i < len; i++) { 19 ma[l++] = s[i]; 20 ma[l++] = '#'; 21 } 22 ma[l] = 0; 23 int mx = 0, id = 0; 24 for (int i = 0; i < l; i++) { 25 mp[i] = mx > i ? min(mp[2*id-i], mx-i) : 1; 26 while (ma[i + mp[i]] == ma[i-mp[i]]) mp[i]++; 27 if (i + mp[i] > mx) { 28 mx = i + mp[i]; 29 id = i; 30 } 31 } 32 long long ans = 0; 33 for (int i = 0; i < 2 * len + 2; i++) { 34 ans += mp[i] / 2; 35 } 36 return ans; 37 } 38 39 int main(void) { 40 ios::sync_with_stdio(false); 41 int T; 42 cin >> T; 43 while (T--) { 44 string s, t; 45 cin >> s >> t; 46 bool flag = true; 47 int left = -1, right = -1; 48 for (auto i = 0; i < s.length(); i++) { 49 if (s[i] != t[i]) { 50 if (flag) { 51 left = i; 52 flag = false; 53 } 54 right = i; 55 } 56 } 57 if (right == -1) { 58 cout << manacher(s) << endl; 59 } else { 60 bool possible = true; 61 for (int i = 0; i <= right - left; i++) { 62 if (s[left + i] != t[right - i]) { 63 possible = false; 64 } 65 } 66 long long ans = 0; 67 if (possible) { 68 ans = 1; 69 while (--left >= 0 && ++right < s.length()) { 70 if (s[left] == s[right] && t[left] == t[right] && s[left] == t[left]) { 71 ans++; 72 } else { 73 break; 74 } 75 } 76 } 77 // with different bytes 78 cout << ans << endl; 79 } 80 } 81 return 0; 82 }
L && M：
看到没有人过就溜了。