【Codeforces Round #540 (Div. 3)】 A B C D1 D2 E F1
2024-05-09 05:46:19
A
题意 给你一升水和两升水的价格 问你怎么买便宜
做法 模拟即可
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
#include <random>
#include <ctime>
using namespace std;#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
//inv[1]=1;
//for(int i=2; i<M; ++i) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
//for(int i=0; i<LIM; ++i) coef[i]=1ll*(P[i]-1)*inv[P[i]]%mod;int main()
{//ios::sync_with_stdio(false);//freopen("a.txt","r",stdin);//freopen("b.txt","w",stdout);int q;scanf("%d",&q);while(q--){long long n,a,b;scanf("%lld%lld%lld",&n,&a,&b);if(n%2==1){if(a*2>=b){printf("%lld\n",n/2*b+a);}else{printf("%lld\n",n*a);}}else{if(a*2>=b){printf("%lld\n",n/2*b);}else{printf("%lld\n",n*a);}}}//fclose(stdin);//fclose(stdout);//cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;return 0;
}B
题意 做法 给你n个东西 每天吃一个 去掉一个 使得偶数天吃的等于奇数天吃的 问你有多少种情况
我们知道去掉一个只是改变奇偶性即可
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
#include <random>
#include <ctime>
using namespace std;#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
//inv[1]=1;
//for(int i=2; i<M; ++i) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
//for(int i=0; i<LIM; ++i) coef[i]=1ll*(P[i]-1)*inv[P[i]]%mod;
const int MAX_N = 200025;
long long sum[MAX_N][2];
long long arr[MAX_N];
int main()
{//ios::sync_with_stdio(false);//freopen("a.txt","r",stdin);//freopen("b.txt","w",stdout);int n,ans = 0;scanf("%d",&n);for(int i = 1;i<=n;++i) scanf("%lld",&arr[i]);for(int i = 1;i<=n;++i)sum[i][(i&1)] = sum[i-1][(i&1)] + arr[i],sum[i][1-(i&1)] = sum[i-1][1-(i&1)];for(int i = 1;i<=n;++i){long long Sum = sum[i-1][0],Sum_ = sum[i-1][1];Sum+=sum[n][1] - sum[i][1];Sum_+=sum[n][0] - sum[i][0];if(Sum==Sum_) ans++;}printf("%d\n",ans);//fclose(stdin);//fclose(stdout);//cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;return 0;
}C
题意构造一个回文矩阵
做法 偶数我们只要填进1/4 即可 奇数我们先补中间列，行 剩下的再补1/4
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
#include <random>
#include <ctime>
using namespace std;#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
//inv[1]=1;
//for(int i=2; i<M; ++i) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
//for(int i=0; i<LIM; ++i) coef[i]=1ll*(P[i]-1)*inv[P[i]]%mod;
int arr[500],ans[25][25],num[1025];
vector<int >vt,vt_,vt__;
int main()
{//ios::sync_with_stdio(false);//freopen("a.txt","r",stdin);//freopen("b.txt","w",stdout);int n;scanf("%d",&n);for(int i = 1;i<=n*n;++i) scanf("%d",&arr[i]),num[arr[i]]++;if(n%2==0){bool flag = true;for(int i = 1;i<=1000;++i){if(num[i]){if(num[i]<4||num[i]%4!=0){flag = false;break;}else vt.push_back(i);}}if(!flag){// cout << "ok " <<endl;printf("NO\n");return 0;}else{int sz = vt.size();int tmp = num[vt[0]]/4,id = 0;for(int i = 1;i<=n/2;i++){for(int j = 1;j<=n/2;j++){if(tmp){ans[i][j] = vt[id];tmp--;}else{id++;tmp = num[vt[id]]/4;ans[i][j] = vt[id];tmp--;}}}for(int i = 1;i<=n/2;i++){for(int j = n/2+1;j<=n;++j){ans[i][j] = ans[i][n-j+1];}}for(int i = n/2+1;i<=n;++i){for(int j = 1;j<=n;++j){ans[i][j] = ans[n-i+1][j];}}printf("YES\n");for(int i = 1;i<=n;++i){for(int j = 1;j<=n;++j){j==n?printf("%d\n",ans[i][j]):printf("%d ",ans[i][j]);}}}}else{bool flag = false;int sum_0=0,sum_1=0;for(int i = 1;i<=1000;++i){if(num[i]){if(num[i]%2==1){sum_1++;}else sum_0++;}}if(sum_1!=1){//cout << "2 ok" << endl;printf("NO\n");return 0;}else{for(int i = 1;i<=1000;++i){if(num[i]%2==1){ans[n/2+1][n/2+1] = i;num[i]--;vt.push_back(i);}else vt.push_back(i);}int sz = vt.size(),check =0,sz_ = 0;for(int i = 0;i<sz;++i){if(num[vt[i]]>=4){check+=num[vt[i]]/4;for(int j = 1;j<=num[vt[i]]/4;j++){if(sz_>=(n/2)*(n/2)){//dbg(sz_);vt__.push_back(vt[i]);vt__.push_back(vt[i]);}else vt_.push_back(vt[i]),sz_++;}if(num[vt[i]]%4==2){vt__.push_back(vt[i]);}}else if(num[vt[i]]==2){vt__.push_back(vt[i]);}}//dbg3(vt_.size(),check,n/2*n/2);if(check<(n/2)*(n/2)){//cout << "3 ok "<<endl;printf("NO\n");return 0;}else{printf("YES\n");int k = 0;for(int i = 1;i<=n/2;i++)ans[n/2+1][i] = vt__[k],k++;for(int i = 1;i<=n/2;++i)ans[i][n/2+1] = vt__[k],k++;int id = 0;for(int i = 1;i<=n/2;i++){for(int j = 1;j<=n/2;j++){ans[i][j] = vt_[id];id++;}}for(int i = n/2+2;i<=n;++i){ans[n/2+1][i] = ans[n/2+1][2*(n/2+1)-i];}for(int i = n/2+2;i<=n;++i){ans[i][n/2+1] = ans[2*(n/2+1)-i][n/2+1];}for(int i = 1;i<=n/2;i++){for(int j = n/2+2;j<=n;++j){ans[i][j] = ans[i][(n/2+1)*2-j];}}for(int i = n/2+2;i<=n;++i){for(int j = 1;j<=n;++j){ans[i][j] = ans[2*(n/2+1)-i][j];}}for(int i = 1;i<=n;++i){for(int j = 1;j<=n;++j){j==n?printf("%d\n",ans[i][j]):printf("%d ",ans[i][j]);}}}}}//fclose(stdin);//fclose(stdout);//cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;return 0;
}D1
D1 和 D2 做法一样 数据范围不同
做法 既然是问我们最小几天 我们可以二分答案 贪心的去取并check即可
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
#include <random>
#include <ctime>
using namespace std;#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
//inv[1]=1;
//for(int i=2; i<M; ++i) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
//for(int i=0; i<LIM; ++i) coef[i]=1ll*(P[i]-1)*inv[P[i]]%mod;
long long arr[200025];
bool cmp(long long a,long long b)
{return a>b;
}
int main()
{//ios::sync_with_stdio(false);//freopen("a.txt","r",stdin);//freopen("b.txt","w",stdout);int n,ans;long long m;long long sum = 0;scanf("%d%lld",&n,&m);for(int i = 1;i<=n;++i) scanf("%lld",&arr[i]),sum+=arr[i];sort(arr+1,arr+1+n,cmp);if(sum<m){printf("-1\n");return 0;}else{int l = 1,r = n;while(l<=r){int k = 0,cnt = 1;long long tmp = 0;int mid = (l+r)>>1;while(cnt<=n){tmp+=max(arr[cnt]-k,1ll*0);if(tmp>=m) break;//dbg3(mid,cnt,k);//dbg2(tmp,arr[cnt]);if(cnt%mid==0){k++;}cnt++;}//dbg2(mid,tmp);if(tmp>=m) ans = mid,r = mid - 1;else l = mid + 1;}printf("%d\n",max(1,ans));}//fclose(stdin);//fclose(stdout);//cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;return 0;
}
E
题目给了一种构造方法 按照他的方法构造即可
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
#include <random>
#include <ctime>
using namespace std;#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
//inv[1]=1;
//for(int i=2; i<M; ++i) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
//for(int i=0; i<LIM; ++i) coef[i]=1ll*(P[i]-1)*inv[P[i]]%mod;int main()
{//ios::sync_with_stdio(false);//freopen("a.txt","r",stdin);//freopen("b.txt","w",stdout);long long n,k;scanf("%lld%lld",&n,&k);if(k*(k-1)<n){printf("NO\n");return 0;}else{int num = n;printf("YES\n");for(int i = 1;i<k;++i){for(int j = i + 1;j<=k;++j){printf("%d %d\n",i,j);num--;if(!num) break;printf("%d %d\n",j,i);num--;if(!num) break;}if(!num) break;}}//fclose(stdin);//fclose(stdout);//cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;return 0;
}F1
做法 简单的dfs判断即可 因为可能你没有2 没有1也要算情况 所以要四个情况集合一起判
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
#include <random>
#include <ctime>
using namespace std;#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
//inv[1]=1;
//for(int i=2; i<M; ++i) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
//for(int i=0; i<LIM; ++i) coef[i]=1ll*(P[i]-1)*inv[P[i]]%mod;
const int MAX_N = 300025;
vector<int > G[300025];
int col[MAX_N],cnt[3],sum[MAX_N][3],ans;
void dfs(int rt,int fa)
{sum[rt][col[rt]]++;int sz = G[rt].size();for(int i = 0;i<sz;++i){int to = G[rt][i];if(to==fa) continue;dfs(to,rt);if((sum[to][1]==0||sum[to][2]==0)&&(cnt[1]-sum[to][1]==0||cnt[2]-sum[to][2]==0)) ans++;sum[rt][0] += sum[to][0];sum[rt][1] += sum[to][1];sum[rt][2] += sum[to][2];}return ;
}
int main()
{//ios::sync_with_stdio(false);//freopen("a.txt","r",stdin);//freopen("b.txt","w",stdout);int n;scanf("%d",&n);for(int i = 1;i<=n;++i) scanf("%d",&col[i]),cnt[col[i]]++;for(int i = 1;i<n;++i){int a,b;scanf("%d%d",&a,&b);G[a].push_back(b);G[b].push_back(a);}dfs(1,-1);printf("%d\n",ans);//fclose(stdin);//fclose(stdout);//cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;return 0;
}【Codeforces Round #540 (Div. 3)】 A B C D1 D2 E F1相关推荐
- 【Codeforces Round #767 (Div. 2)】 C. Meximum Array 题解
[Codeforces Round #767 (Div. 2) ]C. Meximum Array 题解 1629C: Meximum Array 题解 [Codeforces Round #767 ...
- C1. Make Nonzero Sum (easy version)【Codeforces Round #829 (Div. 2】
Codeforces Round #829 (Div. 2)中C1题目 Codeforces比赛记录 文章目录 题目链接: 一.C1. Make Nonzero Sum (easy version) ...
- 【Codeforces Round #548(Div. 2)】Edgy Trees（数学+bfs求连通块）
题目链接 C. Edgy Trees time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- 【Codeforces Round #525(Div. 2)】Ehab and another another xor problem（思维+异或）
题目链接 D. Ehab and another another xor problem time limit per test 1 second memory limit per test 256 ...
- 【Codeforces Round#618 (Div. 2)】C. Anu Has a Function 题解
题目链接:C. Anu Has a Function 题目 Anu has created her own function f: f(x,y)=(x|y)−y where | denotes the ...
- 【Codeforces Round #784 (Div. 4)】【AK题解】
2022年4月30日19:43:21 文章目录 2022年4月30日19:43:21 A. Division? 题目描述 测试样例 题解 B. Triple 题目描述 测试样例 题解 C. Odd/E ...
- 【Codeforces Round #560 (Div. 3)】 A B C D E F1 F2
cf560 A 题意 给你一个 2e5的01串 和 x y要求这个串%10^x = 10 ^ y 那么按照题意模拟 都变成0 那位变成1 即可 /*if you can't see the repay ...
- 【Codeforces Round #547 (Div. 3)】 A B C D E F1 F2 G
A 题意 问你从一个n 到 m 经过多少次数 *2 *3能到m 所以特判一下m能否整除n 然后除了以后根据奇偶去判断求次数 #include <cstdio> #include < ...
- 【Codeforces Round #544 (Div. 3)】 A B C D E F1 F2
这场A是最失误的地方 能int读入 却用char数组读入 感觉失策了 A 题意 给你开始时间和结尾时间 问你这两个时间的中间时间是什么 做法 做法显然 骚操作在此 #include <cstdi ...
最新文章
- java 异常 理解_java中的异常理解
- 完成端口模式下的高性能网络服务器【转】
- MFC中在picture control中显示图像的方法
- （Eclipse打包问题）Export aborted because fatal lint er
- c语言数组转置原理,为什么这个数组转置不对？
- php函数 chm,php函数手册chm最新
- MySQL事务基础，看这篇就够了！
- 26复杂类型比较,使用Compare .NET objects组件
- 问题解决之——未知usb设备设备描述符请求失败（Jlink驱动）
- 在Ubuntu 16.04 64bit上安装谷歌地球Google Earth免费版
- 基于c51的节日彩灯控制器的设计
- Django serializer 方法补充之：depth 与字段生成超链接
- 设计模式之 Factory Mode 工厂模式：Swift 实现
- 【每日一网】Day4：OverFeat流程简单理解
- 新型脑刺激疗法治疗重度抑郁症
- 探路者团队-贪吃蛇(测评人：刘耀泽)
- oracle安装 redo log,Oracle Redo Log 及Logmnr配置使用
- Spring Cloud Alibaba Sentinel - - ＞ 容错机制
- 河南省计算机专业三本排名2015,河南三本大学排名【最新排行榜】
- zookeeper的客户端工具zoolnspector 工具的安装