【Codeforces Round #565 (Div. 3) A B C D E F】解题报告
2024-05-08 02:54:52
Codeforces 565
失踪人口回归
cf 565 A
题意 你可以把一个偶数变成 n/2 , 可以把一个3的倍数变成 n2/3, 可以把一个5的倍数变成 n4/5
问你最少通过多少步可以变成1
解法 我们发现 2/3是最小的 而且生成2这个因子 ，然后 4/5也能生成 2 所以顺序就是 先 3 后 5 再 2
/*if you can't see the repayWhy not just work step by steprubbish is relaxedto ljq
*/
#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>
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;
const ll mod = (int)1e9+7;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);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);/*namespace sgt
{#define mid ((l+r)>>1)#undef mid
}*/int main()
{//ios::sync_with_stdio(false);//freopen("a.txt","r",stdin);//freopen("b.txt","w",stdout);int t;scanf("%d",&t);while(t--){long long n;scanf("%lld",&n);if(n==1){printf("0\n");continue;}int tmp = 0;while(n%3==0&&n!=1){tmp++;n = n/3*2;}while(n%5==0&&n!=1){n = n/5*4;tmp++;}while(n%2==0&&n!=1){n = n / 2;tmp++;}if(n!=1){printf("-1\n");}else printf("%d\n",tmp);}//fclose(stdin);//fclose(stdout);//cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;return 0;
}B
题意 就是给你 n 个数 你能执行若干次两两相交 问你 3 的倍数有多少个 包括 0
那么一拿到数字我们就%=3 然后只需要本身能除尽 然后 1 和 2 ，然后 3个 1 ，3 个 2 统计一下答案即可
/*if you can't see the repayWhy not just work step by steprubbish is relaxedto ljq
*/
#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>
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;
const ll mod = (int)1e9+7;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);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);/*namespace sgt
{#define mid ((l+r)>>1)#undef mid
}*/
const int MAX_N = 1025;
int arr[MAX_N],num[15];
int main()
{//ios::sync_with_stdio(false);//freopen("a.txt","r",stdin);//freopen("b.txt","w",stdout);int t;scanf("%d",&t);while(t--){int n,ans = 0;scanf("%d",&n);num[0] = num[1] = num[2] = 0;for(int i = 1;i<=n;++i) scanf("%d",&arr[i]),arr[i]%=3,num[arr[i]]++;ans = num[0];num[0] = 0;int minn = min(num[1],num[2]);ans += minn;num[1]-=minn;num[2]-=minn;ans+=num[1]/3;num[1]%=3;ans+=num[2]/3;num[2]%=3;printf("%d\n",ans);}//fclose(stdin);//fclose(stdout);//cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;return 0;
}C
一开始读错题意了 没有看见它这6个必须按顺序 于是就写WA了
其实只要关心最后一个数字是否在即可 我这里有个繁琐的做法是用set存进去下标 然后依次比较
如果你只关心能不能凑成 就可以值统计最后一个是否生成
/*if you can't see the repayWhy not just work step by steprubbish is relaxedto ljq
*/
#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>
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;
const ll mod = (int)1e9+7;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);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);/*namespace sgt
{#define mid ((l+r)>>1)#undef mid
}*/
int arr[500025],num[50];
set<int > st[100];
int main()
{//ios::sync_with_stdio(false);//freopen("a.txt","r",stdin);//freopen("b.txt","w",stdout);int n;scanf("%d",&n);int ans = 0;for(int i = 1;i<=n;++i) scanf("%d",&arr[i]),st[arr[i]].insert(i);while(st[4].size()){int it = *st[4].begin();st[4].erase(it);if(st[8].empty()) break;else{set<int> :: iterator tmp = st[8].upper_bound(it);if(tmp==st[8].end()) break;it = *tmp;st[8].erase(*tmp);}if(st[15].empty()) break;else{set<int> :: iterator tmp = st[15].upper_bound(it);if(tmp==st[15].end()) break;it = *tmp;st[15].erase(*tmp);}if(st[16].empty()) break;else{set<int> :: iterator tmp = st[16].upper_bound(it);if(tmp==st[16].end()) break;it = *tmp;st[16].erase(*tmp);}if(st[23].empty()) break;else{set<int> :: iterator tmp = st[23].upper_bound(it);if(tmp==st[23].end()) break;it = *tmp;st[23].erase(*tmp);}if(st[42].empty()) break;else{set<int> :: iterator tmp = st[42].upper_bound(it);if(tmp==st[42].end()) break;it = *tmp;st[42].erase(*tmp);}ans++;}printf("%d\n",n-ans*6);//fclose(stdin);//fclose(stdout);//cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;return 0;
}D
题意 如果一个数是质数 他只能生成质数序列的第他本身个质数 否则会生成他的最大因子
首先我们要知道 肯定是把两个质数分开 然后我们发现 只有大的质数会被小的质数生成 比如 3 是 2 生成的 5 是 3 生成的 那么我们从大到小 找到一个质数就把生成他的质数放在答案序列里面 如果不是质数 那么我们就需要把他放进答案序列里面 消除他最大因子的贡献
一开始忘记 n是 2n re了一发
ans[0]和ans[2]是想要的答案
/*if you can't see the repayWhy not just work step by steprubbish is relaxedto ljq
*/
#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>
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;
const ll mod = (int)1e9+7;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);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);/*namespace sgt
{#define mid ((l+r)>>1)#undef mid
}*/
int arr[400025];
const int MAX_N = 2751150;
int prime[MAX_N];
int flag[MAX_N+5];
int tot,num[MAX_N];
map<int,int > mp;void make_prime()
{for(int i = 2;i<=MAX_N;++i){if(!flag[i]){prime[++tot] = i;mp[i] = tot;for(int j = i + i;j<=MAX_N;j+=i)flag[j] = 1;}}
}
vector<int > ans[4];
bool cmp(int a,int b)
{return a > b;
}
int main()
{//ios::sync_with_stdio(false);//freopen("a.txt","r",stdin);//freopen("b.txt","w",stdout);int n;make_prime();scanf("%d",&n);for(int i = 1;i<=2*n;++i){scanf("%d",&arr[i]);num[arr[i]]++;}int cnt = 0;sort(arr+1,arr+1+2*n,cmp);for(int i = 1;i<=2*n;++i){if(num[arr[i]]){if(!flag[arr[i]]){if(num[mp[arr[i]]]){num[arr[i]]-- ;num[mp[arr[i]]]--;ans[0].push_back(arr[i]);ans[1].push_back(mp[arr[i]]);}else{cnt++;num[arr[i]]--;ans[2].push_back(arr[i]);}}else{cnt++;num[arr[i]]--;ans[2].push_back(arr[i]);for(int j = 2;j*j<=arr[i];++j){if(arr[i]%j==0){num[arr[i]/j]--;ans[3].push_back(arr[i]/j);break;}}}}}if(ans[1].size()){for(int i = 0;i<ans[1].size();++i)printf("%d ",ans[1][i]);}
// for(int i = 0;i<cnt;++i)
// dbg2(i,ans[2][i]);
// dbg(cnt);if(cnt){for(int i = 0;i<cnt;++i)printf("%d ",ans[2][i]);}//fclose(stdin);//fclose(stdout);//cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;return 0;
}E
很简单的染色问题 给点染色即可
/*if you can't see the repayWhy not just work step by steprubbish is relaxedto ljq
*/
#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>
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;
const ll mod = (int)1e9+7;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);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);/*namespace sgt
{#define mid ((l+r)>>1)#undef mid
}*/
vector<int > vt[15],V[200025];
int col[200025];
void dfs(int u,int fa,int COL)
{col[u] = COL;int sz = V[u].size();for(int i = 0;i<sz;++i){int v = V[u][i];if(v==fa) continue;if(col[v]==0){dfs(v,u,3-COL);}}
}
int main()
{//ios::sync_with_stdio(false);//freopen("a.txt","r",stdin);//freopen("b.txt","w",stdout);int t;scanf("%d",&t);while(t--){vector<int > tmp;swap(vt[1],tmp);vector<int > tmp_;swap(vt[2],tmp_);
// dbg2(vt[1].size(),vt[2].size());int n,m,a,b;scanf("%d%d",&n,&m);for(int i = 1;i<=n;++i){vector<int> now;swap(V[i],now);col[i] = 0;}for(int i = 1;i<=m;++i){scanf("%d%d",&a,&b);V[a].push_back(b);V[b].push_back(a);}dfs(1,0,2);for(int i = 1;i<=n;++i)if(col[i]==1) vt[1].push_back(i);else if(col[i]==2) vt[2].push_back(i);int sz = vt[1].size(),sz_ = vt[2].size();
// dbg2(sz,sz_);if(sz>=sz_){printf("%d\n",sz_);for(int i = 0;i<sz_;++i)i==sz_-1?printf("%d\n",vt[2][i]):printf("%d ",vt[2][i]);}else{printf("%d\n",sz);for(int i = 0;i<sz;++i)i==sz-1?printf("%d\n",vt[1][i]):printf("%d ",vt[1][i]);}}//fclose(stdin);//fclose(stdout);//cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;return 0;
}F
题意 就是一个人有 n 回合 每回合不能超过 3 代价的花费 问你能造成多少点伤害
第十个回合是双倍攻击
那么我们就意识到了这个很像dp 而且跟卡牌轮次有关系
所以我们定义dp[n][10] 代表第 n 轮 当前是第几张牌次
我们发现代价为 3 总共只有 6 种花费
1 花费打一张 tmp[1]
2 花费打一张 tmp[2]
3 花费打一张 tmp[3]
1 + 1 2花费打 2 张 tmp[4]
1 + 2 3花费打 2 张 tmp[5]
1 + 1 + 1 3花费打三张 tmp[6]
然后注意dp转移过来的时候合法性 也就是初始化问题
然后这题会爆万恶的int
/*if you can't see the repayWhy not just work step by steprubbish is relaxedto ljq
*/
#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>
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;
const ll mod = (int)1e9+7;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);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);/*namespace sgt
{#define mid ((l+r)>>1)#undef mid
}*/
const int MAX_N = 200025;
long long dp[MAX_N][10];
priority_queue<long long> q[15];
long long k[200025],tmp[15],maxx[15];
int main()
{//ios::sync_with_stdio(false);//freopen("a.txt","r",stdin);//freopen("b.txt","w",stdout);int n;scanf("%d",&n);memset(dp,-1,sizeof(dp));dp[0][0] = 0;for(int i = 1;i<=n;++i){int c ,d;while(!q[1].empty()) q[1].pop();while(!q[2].empty()) q[2].pop();while(!q[3].empty()) q[3].pop();scanf("%d",&k[i]);for(int j = 1;j<=k[i];++j){scanf("%d%d",&c,&d);q[c].push(d);}for(int j = 0;j<10;++j) dp[i][j] = dp[i-1][j];for(int j = 1;j<=6;++j) tmp[j] = -inf;if(!q[1].empty()) tmp[1] = q[1].top();if(!q[2].empty()) tmp[2] = q[2].top();if(!q[3].empty()) tmp[3] = q[3].top();
// cout << "ok" <<endl;if(q[1].size()>=2){long long now = q[1].top();q[1].pop();tmp[4] = q[1].top() + now;maxx[4] = max(q[1].top(),now);q[1].push(now);}
// cout << "OK"<<endl;if(!q[1].empty()&&!q[2].empty()){maxx[5] = max(q[1].top(),q[2].top());tmp[5] = q[1].top()+q[2].top();}
// cout << "ok" << endl;if(q[1].size()>=3){maxx[6] = 0;tmp[6] = q[1].top();maxx[6] = max(maxx[6],q[1].top());q[1].pop();tmp[6] += q[1].top();maxx[6] = max(maxx[6],q[1].top());q[1].pop();tmp[6] += q[1].top();maxx[6] = max(maxx[6],q[1].top());q[1].pop();}
// cout << "ok" <<endl;if(tmp[1]!=-inf){for(int j = 1;j<=9;++j)if(dp[i-1][j-1]!=-1) dp[i][j] = max(dp[i][j],dp[i-1][j-1]+tmp[1]);if(dp[i-1][9]!=-1)dp[i][0] = max(dp[i][0],dp[i-1][9]+tmp[1]*2);}if(tmp[2]!=-inf){for(int j = 1;j<=9;++j)if(dp[i-1][j-1]!=-1) dp[i][j] = max(dp[i][j],dp[i-1][j-1]+tmp[2]);if(dp[i-1][9]!=-1)dp[i][0] = max(dp[i][0],dp[i-1][9]+tmp[2]*2);}if(tmp[3]!=-inf){for(int j = 1;j<=9;++j)if(dp[i-1][j-1]!=-1) dp[i][j] = max(dp[i][j],dp[i-1][j-1]+tmp[3]);if(dp[i-1][9]!=-1)dp[i][0] = max(dp[i][0],dp[i-1][9]+tmp[3]*2);}if(tmp[4]!=-inf){for(int j = 2;j<=9;++j)if(dp[i-1][j-2]!=-1) dp[i][j] = max(dp[i][j],dp[i-1][j-2]+tmp[4]);if(dp[i-1][9]!=-1)dp[i][1] = max(dp[i][1],dp[i-1][9]+tmp[4]+maxx[4]);if(dp[i-1][8]!=-1)dp[i][0] = max(dp[i][0],dp[i-1][8]+tmp[4]+maxx[4]);}if(tmp[5]!=-inf){for(int j = 2;j<=9;++j)if(dp[i-1][j-2]!=-1)dp[i][j] = max(dp[i][j],dp[i-1][j-2]+tmp[5]);if(dp[i-1][9]!=-1)dp[i][1] = max(dp[i][1],dp[i-1][9]+tmp[5]+maxx[5]);if(dp[i-1][8]!=-1)dp[i][0] = max(dp[i][0],dp[i-1][8]+tmp[5]+maxx[5]);}if(tmp[6]!=-inf){for(int j = 3;j<=9;++j)if(dp[i-1][j-3]!=-1)dp[i][j] = max(dp[i][j],dp[i-1][j-3]+tmp[6]);if(dp[i-1][9]!=-1)dp[i][2] = max(dp[i][2],dp[i-1][9]+tmp[6]+maxx[6]);if(dp[i-1][8]!=-1)dp[i][1] = max(dp[i][1],dp[i-1][8]+tmp[6]+maxx[6]);if(dp[i-1][7]!=-1)dp[i][0] = max(dp[i][0],dp[i-1][7]+tmp[6]+maxx[6]);}}long long ans = 0;for(int i = 0;i<10;++i)ans = max(ans,dp[n][i]);printf("%lld\n",ans);//fclose(stdin);//fclose(stdout);//cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;return 0;
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