网站制作工作室,网站建设制作周期,江都区城乡建设局网站马局,营销技巧和话术1、编程模拟
2、栈模拟
3、找规律#xff1f;#xff08;从终止状态思考#xff09;
4、二分
5、找规律#xff0c;数学题
6、贪心#xff08;思维题#xff09;
7、基环树
A - Rook 题意#xff1a; 直接模拟
// Problem: A. Rook
// Contest: Codeforces - C…1、编程模拟
2、栈模拟
3、找规律从终止状态思考
4、二分
5、找规律数学题
6、贪心思维题
7、基环树
A - Rook 题意 直接模拟
// Problem: A. Rook
// Contest: Codeforces - Codeforces Round 913 (Div. 3)
// URL: https://codeforces.com/contest/1907/problem/0
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)#include bits/stdc.h
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second
#define endl \n
const LL maxn 4e057;
const LL N 5e0510;
const LL mod 1e097;
const int inf 0x3f3f3f3f;
const LL llinf 5e18;
typedef pairint,intpl;
priority_queueLL , vectorLL, greaterLL mi;//小根堆
priority_queueLL ma;//大根堆
LL gcd(LL a, LL b){return b 0 ? gcd(b , a % b) : a;
}LL lcm(LL a , LL b){return a / gcd(a , b) * b;
}
int n , m;
vectorinta(N , 0);
void init(int n){for(int i 0 ; i n ; i ){a[i] 0;}
}
void solve()
{string s;cin s;char c s[0];int k s[1] - 0;for(int i 0 ; i 8 ; i){char t i a;if(t c){continue;}else{cout t kendl;}}for(int i 1 ; i 9 ; i ){if(i k)continue;else{cout c i endl;}}
}
int main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout.precision(10);int t1;cint;while(t--){solve();}return 0;
}B - YetnotherrokenKeoard 题意给定键入顺序但是其中‘B’和‘b’不再代表字母‘B’代表删除前面第一个大写字母‘b’代表删除前面第一个小写字母求最终键入的字符串结果。 思路‘第一个’自然想到了栈的后进后出原则‘B’‘b’操作即是出栈操作。先不考虑最终结果而是将其字母的键入时间放入栈当中。最后将栈中键入时间全部取出再根据键入时间输出结果。 // Problem: B. YetnotherrokenKeoard
// Contest: Codeforces - Codeforces Round 913 (Div. 3)
// URL: https://codeforces.com/contest/1907/problem/B
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)#include bits/stdc.h
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second
#define endl \n
const LL maxn 4e057;
const LL N 5e0510;
const LL mod 1e097;
const int inf 0x3f3f3f3f;
const LL llinf 5e18;
typedef pairint,intpl;
priority_queueLL , vectorLL, greaterLL mi;//小根堆
priority_queueLL ma;//大根堆
LL gcd(LL a, LL b){return b 0 ? gcd(b , a % b) : a;
}LL lcm(LL a , LL b){return a / gcd(a , b) * b;
}
int n , m;
vectorinta(N , 0);
void init(int n){for(int i 0 ; i n ; i ){a[i] 0;}
}
void solve()
{string s;cin s;stackintbig , small;int len s.size();for(int i 0 ; i len ; i ){if(s[i] a s[i] z){if(s[i] b){if(!small.empty()){small.pop();}} else{small.push(i);}}else{if(s[i] B){if(!big.empty()){big.pop();}}else{big.push(i);}}}vectorintans;while(!big.empty()){ans.pb(big.top());big.pop();}while(!small.empty()){ans.pb(small.top());small.pop();}sort(ans.begin() , ans.end());len ans.size();for(int i 0 ; i len ; i ){cout s[ans[i]];}cout endl;
}
int main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout.precision(10);int t1;cint;while(t--){solve();}return 0;
}C - Removal of Unattractive Pairs 题意给定字符串能够删除相邻两个不相同字符求最终字符最短能为多少。 思路考虑最终状态一定不存在字符或者只存在一种字符。因此考虑将其余字符全部删除能将字符串变为多短。 // Problem: C. Removal of Unattractive Pairs
// Contest: Codeforces - Codeforces Round 913 (Div. 3)
// URL: https://codeforces.com/contest/1907/problem/C
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)#include bits/stdc.h
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second
#define endl \n
const LL maxn 4e057;
const LL N 5e0510;
const LL mod 1e097;
const int inf 0x3f3f3f3f;
const LL llinf 5e18;
typedef pairint,intpl;
priority_queueLL , vectorLL, greaterLL mi;//小根堆
priority_queueLL ma;//大根堆
LL gcd(LL a, LL b){return b 0 ? gcd(b , a % b) : a;
}LL lcm(LL a , LL b){return a / gcd(a , b) * b;
}
int n , m;
vectorinta(N , 0);
void init(int n){for(int i 0 ; i n ; i ){a[i] 0;}
}
void solve()
{cin n;vectorintcnt(26 , 0);string s;cin s;for(int i 0; i n ; i ){cnt[s[i] - a];}int maxx 0;for(int i 0 ; i 26 ; i ){maxx max(maxx , cnt[i]);}cout max( n 1 ? 1 : 0 , n - 2 * (n - maxx)) endl;
}
int main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout.precision(10);int t1;cint;while(t--){solve();}return 0;
}D - Jumping Through Segments 题意 思路直接二分求答案每一轮的范围是[上一轮左端点 - k 上一轮右端点 k ] ,然后和当前线段合并若不相交则false。 // Problem: D. Jumping Through Segments
// Contest: Codeforces - Codeforces Round 913 (Div. 3)
// URL: https://codeforces.com/contest/1907/problem/D
// Memory Limit: 256 MB
// Time Limit: 5000 ms
//
// Powered by CP Editor (https://cpeditor.org)#include bits/stdc.h
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second
#define endl \n
const LL maxn 4e057;
const LL N 5e0510;
const LL mod 1e097;
const int inf 0x3f3f3f3f;
const LL llinf 5e18;
typedef pairint,intpl;
priority_queueLL , vectorLL, greaterLL mi;//小根堆
priority_queueLL ma;//大根堆
LL gcd(LL a, LL b){return b 0 ? gcd(b , a % b) : a;
}LL lcm(LL a , LL b){return a / gcd(a , b) * b;
}
int n , m;
vectorinta(N , 0);
void init(int n){for(int i 0 ; i n ; i ){a[i] 0;}
}
vectorpairint,intrange;
bool check(int len){pairint,int ran {0 , 0};int t range.size();for(auto to : range){
// cout ran.x ran.y endl;int l to.x;int r to.y;ran.x - len;ran.y len;if(ran.x r || ran.y l)return false;ran.x max(ran.x , l);ran.y min(ran.y , r);}return true;
}
void solve()
{cin n;for(int i 0 ; i n ; i ){pairint,inttmp;cin tmp.x tmp.y;range.pb(tmp);}int l 0 , r 1e9;while(l r){int mid (l r) / 2;if(check(mid)){r mid;} else{l mid 1;}}cout l endl;range.clear();
}
int main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout.precision(10);int t1;cint;while(t--){solve();}return 0;
}E - Good Triples
题意
思路:可以发现若要满足条件那么三个数是不能进位的因此本题中数据的每一位是不会影响的。满足乘法原则将每一位所能带来的贡献相乘即可。
// Problem: E. Good Triples
// Contest: Codeforces - Codeforces Round 913 (Div. 3)
// URL: https://codeforces.com/contest/1907/problem/E
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)#include bits/stdc.h
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second
#define endl \n
const LL maxn 4e057;
const LL N 5e0510;
const LL mod 1e097;
const int inf 0x3f3f3f3f;
const LL llinf 5e18;
typedef pairint,intpl;
priority_queueLL , vectorLL, greaterLL mi;//小根堆
priority_queueLL ma;//大根堆
LL gcd(LL a, LL b){return b 0 ? gcd(b , a % b) : a;
}LL lcm(LL a , LL b){return a / gcd(a , b) * b;
}
int n , m;
vectorinta(N , 0);
void init(int n){for(int i 0 ; i n ; i ){a[i] 0;}
}
int dig(int s){int cnt 0;while(s){cnt s % 10;s/10;}return cnt;
}
void solve()
{// 0 1 // 1 3// 2 6// 3 10// 4 15// 5 21// 6 28// 7 36// 8 45// 9 55// 3141 3 * 15 * 3 * 10LL mask[10] {1 , 3 , 6 , 10 , 15 , 21 , 28 , 36 , 45 , 55};string s;cin s;int len s.size();LL ans 1;for(int i 0 ; i len ; i ){ans * mask[s[i] - 0];}cout ans endl;
}
int main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout.precision(10);int t1;cint;while(t--){solve();}return 0;
}F - Shift and Reverse 思路将数组放到环上考虑 假设环起初按顺时针排 且有一个指针指向了环的第一位。现考虑操作变成了什么
移位操作将指针往前移动一格。
反转将环变成逆时针将非递减转化为非递增。
因此若要满足非递减排序必然存在从环上某一点出发顺时针是非递减顺序/非递增的。如何去模拟环只需要将整个数组复制一遍到尾端即可模拟。然后就是找非递减序列和非递增序列即可。然后判断一下需要多少操作。 // Problem: F. Shift and Reverse
// Contest: Codeforces - Codeforces Round 913 (Div. 3)
// URL: https://codeforces.com/contest/1907/problem/F
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)#include bits/stdc.h
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second
#define endl \n
const LL maxn 4e057;
const LL N 5e0510;
const LL mod 1e097;
const int inf 0x3f3f3f3f;
const LL llinf 5e18;
typedef pairint,intpl;
priority_queueLL , vectorLL, greaterLL mi;//小根堆
priority_queueLL ma;//大根堆
LL gcd(LL a, LL b){return b 0 ? gcd(b , a % b) : a;
}LL lcm(LL a , LL b){return a / gcd(a , b) * b;
}
int n , m;
vectorinta(N , 0);
void init(int n){for(int i 0 ; i n ; i ){a[i] 0;}
}
void solve()
{cin n;for(int i 0 ; i n ; i )cin a[i];vectorintv(2 * n , 0);for(int i 0 ; i n ; i ){v[i] a[i];v[i n] a[i];}int ans -1;//递减for(int i 0 ; i n ; ){int st i;int cnt 1;while(i 2 * n - 1 v[i] v[i 1]){cnt;i;}if(cnt n){ans min(1 st, 1 (n - st));}i ;}//递增for(int i 0 ; i n ; ){int st i;int cnt 1;while(i 2 * n - 1 v[i] v[i 1]){cnt;i;}if(cnt n){if(ans -1){ans min(n - st , st 2);}else{ans min(ans ,min(n - st , st 2) );}if(st 0)ans 0;}i ;} cout ans endl;
}
int main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout.precision(10);int t1;cint;while(t--){solve();}return 0;
}G - Lights 题意 思路将相关联的灯建边每次操作都对边两端处理。可以发现这是一颗基环树也就是说在非环上的点可以按照拓扑序来唯一的去处理要么1变0同时另一端也变化要么不变。现在就只剩下环上的一部分了可以发现若环上剩余亮着的灯是奇数的话那么就不能全部熄灭了。 接下来考虑输出最小操作若环上某一条边的操作已经确定下来了变或者不变那么其余所有边操作都确定了因此可以考虑类似于环形dp的思想。将所有边操作分为两类表示对一类中的每一条边操作后整个环都能变成0。最后比较两类边集大小考虑输出哪一类即可。 // Problem: G. Lights
// Contest: Codeforces - Codeforces Round 913 (Div. 3)
// URL: https://codeforces.com/contest/1907/problem/G
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)#include bits/stdc.h
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second
#define endl \n
const LL maxn 4e057;
const LL N 5e0510;
const LL mod 1e097;
const int inf 0x3f3f3f3f;
const LL llinf 5e18;
typedef pairint,intpl;
priority_queueLL , vectorLL, greaterLL mi;//小根堆
priority_queueLL ma;//大根堆
LL gcd(LL a, LL b){return b 0 ? gcd(b , a % b) : a;
}LL lcm(LL a , LL b){return a / gcd(a , b) * b;
}
int n , m;
vectorinta(N , 0);
void init(int n){for(int i 0 ; i n ; i ){a[i] 0;}
}
void solve()
{cin n; vectorintin(n 5 , 0);vectorintvis(n 5 , 0);vectorinte(n 5 , 0) ;string s;cin s;s s;for(int i 1 ; i n ; i ){int u i , v;cin v;e[u] v;in[v];} queueintq;for(int i 1 ; i n ;i ){if(in[i] 0){q.push(i);}}vectorintans;while(!q.empty()){int x q.front();q.pop();if(s[x] 1){s[x] 0;ans.pb(x);if(s[e[x]] 1){s[e[x]] 0;}else{s[e[x]] 1;}}in[e[x]]--;if(in[e[x]] 0){q.push(e[x]);}}for(int i 1 ; i n ; i ){if(in[i]){int j i;int t 0;//分为两组int len 0;int res 0;//t 1 的组需要操作的数while(in[j]){if(s[j] 1){t ^ 1;}res t;in[j] 0;len ;j e[j];}if(t 1){cout -1\n;return;}for(int k 0 ; k len ; k ){if(s[j] 1){t ^ 1;}if(t (res len - res)){ans.pb(j);}j e[j];}}}cout ans.size() endl;for(auto it : ans){cout it ;}
}
int main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout.precision(10);int t1;cint;while(t--){solve();}return 0;
}
