A题,题意是给一个回文串,问有多少个字符删掉,还是一个回文串
这个题看样例,肯定是从中间开始查相同字符的段长度,没啥难度
代码:
#include <bits/stdc++.h> using namespace std; constexpr int limit = (3000000 + 5);//防止溢出 #define INF 0x3f3f3f3f #define inf 0x3f3f3f3f3f #define lowbit(i) i&(-i)//一步两步 #define EPS 1e-9 #define FASTIO ios::sync_with_stdio(false);cin.tie(0),cout.tie(0); #define ff(a) printf("%d\n",a ); #define pi(a, b) pair<a,b> #define rep(i, a, b) for(ll i = a; i <= b ; ++i) #define per(i, a, b) for(ll i = b ; i >= a ; --i) #define MOD 998244353 #define traverse(u) for(int i = head[u]; ~i ; i = edge[i].nxt) #define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin) #define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout) typedef long long ll; typedef unsigned long long ull; char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf; inline ll read() { #define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++) ll sign = 1, x = 0; char s = getchar(); while (s > '9' || s < '0') { if (s == '-')sign = -1; s = getchar(); } while (s >= '0' && s <= '9') { x = (x << 3) + (x << 1) + s - '0'; s = getchar(); } return x * sign; #undef getchar }//快读 void print(ll x) { if (x / 10) print(x / 10); *O++ = x % 10 + '0'; } void write(ll x, char c = 't') { if (x < 0)putchar('-'), x = -x; print(x); if (!isalpha(c))*O++ = c; fwrite(obuf, O - obuf, 1, stdout); O = obuf; } int n,m; int a[limit]; int b[limit]; void solve(){ cin>>n; string str; cin>>str; str = " " + str; int ans = 0; int l = (n + 1) / 2; per(i, 1, l){ if(str[i] != str[l]){ break; } ++ans; } rep(i, l + 1, n){ if(str[i] != str[l]){ break; } ++ans; } cout<<ans<<endl; }; int32_t main() { #ifdef LOCAL FOPEN; // FOUT; #endif FASTIO int kase; cin>>kase; while (kase--) solve(); cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s\n"; return 0; }AC Code
B题,题意是选取一个非负整数X,当p[i] & p[j] == X,可以交换i和j,然后在任意次交换位置之后让permutation p 有序
如果有很多X输出最大的X
这个题属实没啥难度,很容易联想到交换位置的话,所有不在排序后自己位置上的元素都需要swap,然后通过某种冒泡排序,就能交换到有序
那么这个X一定是被所有的不在位置上的元素share的,所以把他们串起来取and就行了
#include <bits/stdc++.h> using namespace std; constexpr int limit = (3000000 + 5);//防止溢出 #define INF 0x3f3f3f3f #define inf 0x3f3f3f3f3f #define lowbit(i) i&(-i)//一步两步 #define EPS 1e-9 #define FASTIO ios::sync_with_stdio(false);cin.tie(0),cout.tie(0); #define ff(a) printf("%d\n",a ); #define pi(a, b) pair<a,b> #define rep(i, a, b) for(ll i = a; i <= b ; ++i) #define per(i, a, b) for(ll i = b ; i >= a ; --i) #define MOD 998244353 #define traverse(u) for(int i = head[u]; ~i ; i = edge[i].nxt) #define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin) #define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout) typedef long long ll; typedef unsigned long long ull; char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf; inline ll read() { #define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++) ll sign = 1, x = 0; char s = getchar(); while (s > '9' || s < '0') { if (s == '-')sign = -1; s = getchar(); } while (s >= '0' && s <= '9') { x = (x << 3) + (x << 1) + s - '0'; s = getchar(); } return x * sign; #undef getchar }//快读 void print(ll x) { if (x / 10) print(x / 10); *O++ = x % 10 + '0'; } void write(ll x, char c = 't') { if (x < 0)putchar('-'), x = -x; print(x); if (!isalpha(c))*O++ = c; fwrite(obuf, O - obuf, 1, stdout); O = obuf; } int n,m; int a[limit]; int b[limit]; void solve(){ cin>>n; rep(i,1,n){ cin>>a[i]; b[i] = a[i]; } sort(b + 1, b + 1 + n); ll ans = LONG_LONG_MAX; rep(i,1,n){ if(b[i] != a[i]){ ans &= a[i]; } } cout<<ans<<endl; }; int32_t main() { #ifdef LOCAL FOPEN; // FOUT; #endif FASTIO int kase; cin>>kase; while (kase--) solve(); cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s\n"; return 0; }AC Code
C题,给一个序列,你可以任意安排这个序列的所有元素顺序,使得min(a, reverse(a))的LIS长度最小值最大
然后很容易想到那就各取一半,然后首先每个不重复的元素对答案的贡献是1/2,因为可以被加到前后的LIS里面,然后每个重复的元素对答案贡献是1
简单题,我怎么感觉我还没蓝之前做过这个?
#include <bits/stdc++.h> using namespace std; constexpr int limit = (3000000 + 5);//防止溢出 #define INF 0x3f3f3f3f #define inf 0x3f3f3f3f3f #define lowbit(i) i&(-i)//一步两步 #define EPS 1e-9 #define FASTIO ios::sync_with_stdio(false);cin.tie(0),cout.tie(0); #define ff(a) printf("%d\n",a ); #define pi(a, b) pair<a,b> #define rep(i, a, b) for(ll i = a; i <= b ; ++i) #define per(i, a, b) for(ll i = b ; i >= a ; --i) #define MOD 998244353 #define traverse(u) for(int i = head[u]; ~i ; i = edge[i].nxt) #define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin) #define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout) typedef long long ll; typedef unsigned long long ull; char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf; inline ll read() { #define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++) ll sign = 1, x = 0; char s = getchar(); while (s > '9' || s < '0') { if (s == '-')sign = -1; s = getchar(); } while (s >= '0' && s <= '9') { x = (x << 3) + (x << 1) + s - '0'; s = getchar(); } return x * sign; #undef getchar }//快读 void print(ll x) { if (x / 10) print(x / 10); *O++ = x % 10 + '0'; } void write(ll x, char c = 't') { if (x < 0)putchar('-'), x = -x; print(x); if (!isalpha(c))*O++ = c; fwrite(obuf, O - obuf, 1, stdout); O = obuf; } int n,m; int a[limit]; int b[limit]; void solve(){ cin>>n; map<int, int>mp; rep(i,1,n){ cin>>a[i]; mp[a[i]]++; } int tot = 0; int ans = 0, delta = 0; for(auto [k, v] : mp){ if(v == 1){ ++delta; }else if(v >= 2){ ++ans; } } cout<<ans + (delta + 1) / 2<<endl; }; int32_t main() { #ifdef LOCAL FOPEN; // FOUT; #endif FASTIO int kase; cin>>kase; while (kase--) solve(); cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s\n"; return 0; }AC Code
D题,D真的是个喵喵题,给出一个01串,n个点,顺时针排成圆形,每个位置上0或者1代表这个点的deg是奇数或者偶数,让你构造一棵树,使得
1. 每个点度数奇偶性满足要求
2. 不能有遍的重叠
或者返回这棵树根本不存在
首先很容易想到的有几个很明显的点
1. 树的度数一定为偶数(n - 1条边,每条边对度数的共线为2)
2. 奇数个数节点也必然是偶数
3. 奇数的节点可以为叶子节点,但偶数的节点一定不是叶子节点
然后想了半晌,emmmm,没辙,脑子里只有一个很模糊的想法,就是先找到偶数节点前后一致的点作为树根,然后乱搞,遂g
想了40分钟之后决定看看答案,答案其实很妙,因为看到了第一句话我就会了
就是将01串贪心处理成所有
regex: 1(0)*的链的形式,然后把所有的链处理出来,连边
最后我们会剩下至少两个链/点,然后把一个长度大于1的链的偶数节点当作树根,然后把其他链的最后一个偶数节点链接上去
这样因为我们必定有偶数个链(偶数个1),所以那个树根的度数也必然是偶数
nice!
然后要注意一下,比如n个1的情况,这个就把剩下的n - 1个点(奇数个)连到1上就行了,特判一下
#include <bits/stdc++.h> using namespace std; constexpr int limit = (3000000 + 5);//防止溢出 #define INF 0x3f3f3f3f #define inf 0x3f3f3f3f3f #define lowbit(i) i&(-i)//一步两步 #define EPS 1e-9 #define FASTIO ios::sync_with_stdio(false);cin.tie(0),cout.tie(0); #define ff(a) printf("%d\n",a ); #define pi(a, b) pair<a,b> #define rep(i, a, b) for(ll i = a; i <= b ; ++i) #define per(i, a, b) for(ll i = b ; i >= a ; --i) #define MOD 998244353 #define traverse(u) for(int i = head[u]; ~i ; i = edge[i].nxt) #define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin) #define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout) typedef long long ll; typedef unsigned long long ull; char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf; inline ll read() { #define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++) ll sign = 1, x = 0; char s = getchar(); while (s > '9' || s < '0') { if (s == '-')sign = -1; s = getchar(); } while (s >= '0' && s <= '9') { x = (x << 3) + (x << 1) + s - '0'; s = getchar(); } return x * sign; #undef getchar }//快读 void print(ll x) { if (x / 10) print(x / 10); *O++ = x % 10 + '0'; } void write(ll x, char c = 't') { if (x < 0)putchar('-'), x = -x; print(x); if (!isalpha(c))*O++ = c; fwrite(obuf, O - obuf, 1, stdout); O = obuf; } int n,m; int a[limit]; int b[limit]; int deg[limit]; void solve(){ cin>>n; string str; cin>>str; str = " " + str; int total = 0; if((total = count(str.begin(),str.end(),'1') ) & 1){ cout<<"No"<<endl; return; } if(!total){ cout<<"No"<<endl; return; } vector<pi(int, int)>ans; auto nxt = [&](int x)->int{ ++x; if(x > n) x = 1; return x; }; vector<pi(int, int)>chain; int appender = -1; for(int i = 1, tot = 1; tot <= n; ){ if(str[i] == '1'){ chain.push_back({i, i}); ++tot; int last = i; i = nxt(i); while(tot <= n and str[i] == '0'){ chain.back().second = i; ++tot; ans.push_back({last, i}); last = i; // cout<<i<<" "<<last<<endl; i = nxt(i); } auto [l, r] = chain.back(); // cout<<l<<" "<<r<<endl; if(l != r and appender == -1){ appender = last; chain.pop_back(); } } else{ i = nxt(i); } } if(appender == -1){ ans.clear(); rep(i,2,n){ ans.push_back({1, i}); } }else{ for(auto [l, r] : chain){ ans.push_back({appender, r}); } } cout<<"Yes"<<endl; for(auto [l, r] : ans){ cout<<l<<" "<<r<<endl; } }; int32_t main() { #ifdef LOCAL FOPEN; // FOUT; #endif FASTIO int kase; cin>>kase; while (kase--) solve(); cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s\n"; return 0; }AC Code
能花时间细细想想问题还是对自己思维有帮助的,要不然思维完全就退化了,这很致命,尤其是在当下的寒冬,不能坐在ICPC和队友的功劳簿上吃一辈子!
标签:793,int,ll,Codeforces,long,偶数,str,div2,define From: https://www.cnblogs.com/tiany7/p/16984182.html