A题 给出n个数，问这n个数能不能分成奇数个连续的长度为奇数并且首尾均为奇数的序列 Codeforces849A 题解传送门 代码 1 #include bits/stdc++.h 2 #define ll long long 3 #define ull unsigned long long 4 #de
A题
给出n个数，问这n个数能不能分成奇数个连续的长度为奇数并且首尾均为奇数的序列
Codeforces849A
题解传送门
代码
1 #include <bits/stdc++.h> 2 #define ll long long 3 #define ull unsigned long long 4 #define ms(a,b) memset(a,b,sizeof(a)) 5 const int inf=0x3f3f3f3f; 6 const ll INF=0x3f3f3f3f3f3f3f3f; 7 const int maxn=1e6+10; 8 const int mod=1e9+7; 9 const int maxm=1e3+10; 10 using namespace std; 11 int a[maxm]; 12 int main(int argc, char const *argv[]) 13 { 14 #ifndef ONLINE_JUDGE 15 freopen("/home/wzy/in.txt", "r", stdin); 16 freopen("/home/wzy/out.txt", "w", stdout); 17 srand((unsigned int)time(NULL)); 18 #endif 19 ios::sync_with_stdio(false); 20 cin.tie(0); 21 int n; 22 cin>>n; 23 int sum=0; 24 for(int i=0;i<n;i++) 25 { 26 cin>>a[i]; 27 a[i]&=1; 28 sum+=a[i]; 29 } 30 if(!(n&1)) 31 { 32 cout<<"No\n"; 33 return 0; 34 } 35 if(!a[0]||!a[n-1]) 36 { 37 cout<<"No\n"; 38 return 0; 39 } 40 cout<<"Yes\n"; 41 #ifndef ONLINE_JUDGE 42 cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl; 43 #endif 44 return 0; 45 }View Code
B题
Codeforces872A
注意第一行和第二行出现同一个数的情况，记录一下
代码
1 #include <bits/stdc++.h> 2 #define ll long long 3 #define ull unsigned long long 4 #define ms(a,b) memset(a,b,sizeof(a)) 5 const int inf=0x3f3f3f3f; 6 const ll INF=0x3f3f3f3f3f3f3f3f; 7 const int maxn=1e6+10; 8 const int mod=1e9+7; 9 const int maxm=1e3+10; 10 using namespace std; 11 int a[maxn]; 12 int b[maxn]; 13 int vis[maxn]; 14 int main(int argc, char const *argv[]) 15 { 16 #ifndef ONLINE_JUDGE 17 freopen("/home/wzy/in.txt", "r", stdin); 18 freopen("/home/wzy/out.txt", "w", stdout); 19 srand((unsigned int)time(NULL)); 20 #endif 21 ios::sync_with_stdio(false); 22 cin.tie(0); 23 int n,m; 24 cin>>n>>m; 25 for(int i=0;i<n;i++) 26 { 27 cin>>a[i]; 28 vis[a[i]]=1; 29 } 30 for(int i=0;i<m;i++) 31 cin>>b[i]; 32 sort(a,a+n); 33 sort(b,b+m); 34 int flag=0; 35 for(int i=0;i<m;i++) 36 { 37 if(vis[b[i]]) 38 { 39 cout<<b[i]<<endl; 40 flag=1; 41 break; 42 } 43 } 44 if(!flag) 45 cout<<min(a[0],b[0])<<max(a[0],b[0])<<endl; 46 #ifndef ONLINE_JUDGE 47 cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl; 48 #endif 49 return 0; 50 }View Code
C题
输出四行1.00
D题
不会
E题
给出一个有n个整数的数组 a1, a2, ..., an 和一个整数k。你被要求把这个数组分成k 个非空的子段。 然后从每个k 个子段拿出最小值，再从这些最小值中拿出最大值。求这个最大值最大能为多少？
Codeforces872B
题解传送门
代码
1 #include <bits/stdc++.h> 2 #define ll long long 3 #define ull unsigned long long 4 #define ms(a,b) memset(a,b,sizeof(a)) 5 const int inf=0x3f3f3f3f; 6 const ll INF=0x3f3f3f3f3f3f3f3f; 7 const int maxn=1e6+10; 8 const int mod=1e9+7; 9 const int maxm=1e3+10; 10 using namespace std; 11 int a[maxn]; 12 int main(int argc, char const *argv[]) 13 { 14 #ifndef ONLINE_JUDGE 15 freopen("/home/wzy/in.txt", "r", stdin); 16 freopen("/home/wzy/out.txt", "w", stdout); 17 srand((unsigned int)time(NULL)); 18 #endif 19 ios::sync_with_stdio(false); 20 cin.tie(0); 21 int n,k; 22 cin>>n>>k; 23 int maxx; 24 int place=0; 25 int minn; 26 for(int i=0;i<n;i++) 27 { 28 cin>>a[i]; 29 if(!i) 30 { 31 maxx=a[i]; 32 place=0; 33 minn=a[i]; 34 } 35 else if(maxx<a[i]) 36 place=i,maxx=a[i]; 37 minn=min(minn,a[i]); 38 } 39 if(k==1) 40 cout<<minn<<endl; 41 else if(k==2) 42 cout<<max(a[0],a[n-1])<<endl; 43 else 44 cout<<maxx<<endl; 45 #ifndef ONLINE_JUDGE 46 cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl; 47 #endif 48 return 0; 49 }View Code
F题
给出一个有n个数的序列a[1],a[2]……a[n]，使得在i<=j<=k的条件下，令p*a[i]+q*a[j]+r*a[k]的值最大并输出这个值
Codeforces855D
题解传送门
代码
1 #include <bits/stdc++.h> 2 #define ll long long 3 #define ull unsigned long long 4 #define ms(a,b) memset(a,b,sizeof(a)) 5 const int inf=0x3f3f3f3f; 6 const ll INF=0x3f3f3f3f3f3f3f3f; 7 const int maxn=1e6+10; 8 const int mod=1e9+7; 9 const int maxm=1e3+10; 10 using namespace std; 11 ll Left[maxn]; 12 ll Right[maxn]; 13 ll a[maxn]; 14 int main(int argc, char const *argv[]) 15 { 16 #ifndef ONLINE_JUDGE 17 freopen("/home/wzy/in.txt", "r", stdin); 18 freopen("/home/wzy/out.txt", "w", stdout); 19 srand((unsigned int)time(NULL)); 20 #endif 21 ios::sync_with_stdio(false); 22 cin.tie(0); 23 int n; 24 ll p,q,r; 25 cin>>n>>p>>q>>r; 26 for(int i=0;i<n;i++) 27 cin>>a[i]; 28 Left[0]=a[0]*p; 29 Right[n-1]=a[n-1]*r; 30 for(int i=1;i<n;i++) 31 Left[i]=max(Left[i-1],p*a[i]); 32 for(int i=n-2;i>=0;i--) 33 Right[i]=max(Right[i+1],r*a[i]); 34 ll ans=-INF; 35 for(int i=0;i<n;i++) 36 ans=max(ans,Left[i]+q*a[i]+Right[i]); 37 cout<<ans<<endl; 38 #ifndef ONLINE_JUDGE 39 cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl; 40 #endif 41 return 0; 42 }View Code
G题
排下序就好了
HDU6292
代码
1 #include <bits/stdc++.h> 2 #define ll long long 3 #define ull unsigned long long 4 #define ms(a,b) memset(a,b,sizeof(a)) 5 const int inf=0x3f3f3f3f; 6 const ll INF=0x3f3f3f3f3f3f3f3f; 7 const int maxn=1e6+10; 8 const int mod=1e9+7; 9 const int maxm=1e3+10; 10 using namespace std; 11 int a[maxn]; 12 int b[maxn]; 13 int main(int argc, char const *argv[]) 14 { 15 #ifndef ONLINE_JUDGE 16 freopen("/home/wzy/in.txt", "r", stdin); 17 freopen("/home/wzy/out.txt", "w", stdout); 18 srand((unsigned int)time(NULL)); 19 #endif 20 ios::sync_with_stdio(false); 21 cin.tie(0); 22 int t; 23 int _=0; 24 cin>>t; 25 while(t--) 26 { 27 int n,m; 28 cin>>n>>m; 29 for(int i=0;i<n;i++) 30 cin>>a[i]; 31 for(int i=0;i<m;i++) 32 cin>>b[i]; 33 sort(a,a+n); 34 sort(b,b+m); 35 cout<<"Problem "<<1000+(++_)<<":"<<endl; 36 if(!n) 37 cout<<"Shortest judge solution: N/A bytes."<<endl; 38 else 39 cout<<"Shortest judge solution: "<<a[0]<<" bytes."<<endl; 40 if(!m) 41 cout<<"Shortest team solution: N/A bytes."<<endl; 42 else 43 cout<<"Shortest team solution: "<<b[0]<<" bytes."<<endl; 44 } 45 #ifndef ONLINE_JUDGE 46 cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl; 47 #endif 48 return 0; 49 }View Code
H题
现在有n个整数，在这n个数中找出k个数，保证这k个数中任意两个数差的绝对值可以被m整除。
Codeforces876B
题解传送门
代码
1 #include <bits/stdc++.h> 2 #define ll long long 3 #define ull unsigned long long 4 #define ms(a,b) memset(a,b,sizeof(a)) 5 const int inf=0x3f3f3f3f; 6 const ll INF=0x3f3f3f3f3f3f3f3f; 7 const int maxn=1e6+10; 8 const int mod=1e9+7; 9 const int maxm=1e3+10; 10 using namespace std; 11 int a[maxn]; 12 int vis[maxn]; 13 bool cmp(int a,int b) 14 { 15 return a>b; 16 } 17 int main(int argc, char const *argv[]) 18 { 19 #ifndef ONLINE_JUDGE 20 freopen("/home/wzy/in.txt", "r", stdin); 21 freopen("/home/wzy/out.txt", "w", stdout); 22 srand((unsigned int)time(NULL)); 23 #endif 24 ios::sync_with_stdio(false); 25 cin.tie(0); 26 int n,m,k; 27 cin>>n>>k>>m; 28 int cnt=0; 29 int num; 30 int maxx=0; 31 for(int i=0;i<n;i++) 32 { 33 cin>>a[i]; 34 if(!vis[a[i]%m]) 35 cnt++; 36 vis[a[i]%m]++; 37 if(maxx<vis[a[i]%m]) 38 maxx=vis[a[i]%m],num=a[i]%m; 39 } 40 if(maxx<k) 41 cout<<"No\n"; 42 else 43 { 44 int res=0; 45 cout<<"Yes\n"; 46 for(int i=0;i<n;i++) 47 { 48 if(a[i]%m==num) 49 { 50 res++; 51 cout<<a[i]<<" "; 52 } 53 if(res==k) 54 break; 55 } 56 cout<<"\n"; 57 } 58 #ifndef ONLINE_JUDGE 59 cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl; 60 #endif 61 return 0; 62 }View Code
I题
HDU5965
题解传送门
先固定第一个位置的雷的数量，往后递推
代码
1 #include <bits/stdc++.h> 2 #define ll long long 3 #define ull unsigned long long 4 #define ms(a,b) memset(a,b,sizeof(a)) 5 const int inf=0x3f3f3f3f; 6 const ll INF=0x3f3f3f3f3f3f3f3f; 7 const int maxn=1e4+10; 8 const ll mod=1e8+7; 9 const int maxm=1e3+10; 10 using namespace std; 11 ll a[maxn]; 12 // 每一列可以放多少个 13 ll dp[maxn]; 14 int main(int argc, char const *argv[]) 15 { 16 #ifndef ONLINE_JUDGE 17 freopen("/home/wzy/in.txt", "r", stdin); 18 freopen("/home/wzy/out.txt", "w", stdout); 19 srand((unsigned int)time(NULL)); 20 #endif 21 ios::sync_with_stdio(false); 22 cin.tie(0); 23 int t; 24 cin>>t; 25 while(t--) 26 { 27 ms(dp,0); 28 ms(a,0); 29 string s; 30 cin>>s; 31 int l=s.length(); 32 for(int i=0;i<l;i++) 33 a[i]=1LL*(s[i]-‘0‘); 34 ll ans=0; 35 // 固定第一个位置的个数 36 for(int i=0;i<3;i++) 37 { 38 if(i>a[0]) 39 break; 40 dp[0]=i; 41 ll pos=1; 42 // 第一个固定之后，枚举后面的每个位置 43 int cnt=0; 44 for(int j=1;j<l;j++) 45 { 46 int res; 47 // 第一列前面没有别的了，所以不需要考虑第一列左边的情况 48 if(j==1) 49 res=a[j-1]-dp[j-1]; 50 // 第j列需要放的雷的个数 51 else 52 res=a[j-1]-dp[j-1]-dp[j-2]; 53 // 符合要求 54 if(res<=2&&res>=0) 55 dp[j]=res,cnt++; 56 } 57 // 如果没有枚举到最后一个位置 58 if(cnt!=l-1) 59 continue; 60 // 如果最后两列放雷数不等于实际的个数，排除掉 61 if(dp[l-1]+dp[l-2]!=a[l-1]) 62 continue; 63 // 计算当第一列为雷数为i的总情况 64 for(int j=0;j<l;j++) 65 if(dp[j]==1) 66 pos<<=1,pos%=mod; 67 ans+=pos;ans%=mod; 68 } 69 cout<<ans%mod<<endl; 70 } 71 #ifndef ONLINE_JUDGE 72 cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl; 73 #endif 74 return 0; 75 }View Code
J题
Codeforces913C
题解传送门
注意考虑两种情况
代码
1 #include <bits/stdc++.h> 2 #define ll long long 3 #define ull unsigned long long 4 #define ms(a,b) memset(a,b,sizeof(a)) 5 const int inf=0x3f3f3f3f; 6 const ll INF=0x3f3f3f3f3f3f3f3f; 7 const int maxn=1e6+10; 8 const int mod=1e9+7; 9 const int maxm=1e3+10; 10 using namespace std; 11 struct wzy 12 { 13 ll bigness; 14 ll money; 15 double price; 16 }p[maxn]; 17 bool cmp(wzy u,wzy v) 18 { 19 return u.price<v.price; 20 } 21 int main(int argc, char const *argv[]) 22 { 23 #ifndef ONLINE_JUDGE 24 freopen("/home/wzy/in.txt", "r", stdin); 25 freopen("/home/wzy/out.txt", "w", stdout); 26 srand((unsigned int)time(NULL)); 27 #endif 28 ios::sync_with_stdio(false); 29 cin.tie(0); 30 int n; 31 ll l; 32 cin>>n>>l; 33 for(int i=1;i<=n;i++) 34 { 35 cin>>p[i].money; 36 p[i].bigness=(1LL<<(i-1)); 37 p[i].price=1.0*p[i].money/p[i].bigness; 38 } 39 sort(p+1,p+1+n,cmp); 40 ll L=l; 41 ll ans=INF; 42 ll res1=0,res2=0; 43 while(L>0) 44 { 45 for(int i=1;i<=n;i++) 46 { 47 // 全买这个饮料 48 res1=res2+ceil(1.0*L/p[i].bigness)*p[i].money; 49 ans=min(ans,res1); 50 // 多余的部分买别的 51 res2+=L/p[i].bigness*p[i].money; 52 L-=(p[i].bigness*(L/p[i].bigness)); 53 } 54 } 55 ans=min(ans,res2); 56 cout<<ans<<endl; 57 #ifndef ONLINE_JUDGE 58 cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl; 59 #endif 60 return 0; 61 }View Code
K题
矩阵快速幂
f[i]=f[i-1]+2*f[i-2]+n^3
HDU6470
题解传送门
L题
暴力枚举就行了
Codeforces879A
代码
1 #include <bits/stdc++.h> 2 #define ll long long 3 #define ull unsigned long long 4 #define ms(a,b) memset(a,b,sizeof(a)) 5 const int inf=0x3f3f3f3f; 6 const ll INF=0x3f3f3f3f3f3f3f3f; 7 const int maxn=1e6+10; 8 const int mod=1e9+7; 9 const int maxm=1e3+10; 10 using namespace std; 11 struct wzy 12 { 13 ll s,d; 14 }p[maxn]; 15 int main(int argc, char const *argv[]) 16 { 17 #ifndef ONLINE_JUDGE 18 freopen("/home/wzy/in.txt", "r", stdin); 19 freopen("/home/wzy/out.txt", "w", stdout); 20 srand((unsigned int)time(NULL)); 21 #endif 22 ios::sync_with_stdio(false); 23 cin.tie(0); 24 int n; 25 cin>>n; 26 for(int i=0;i<n;i++) 27 cin>>p[i].d>>p[i].s; 28 ll ans=p[0].d; 29 for(int i=1;i<n;i++) 30 { 31 ans+=1; 32 ll res=p[i].d; 33 if(res>=ans) 34 ans=res; 35 else 36 { 37 ll pos=ceil(1.0*(ans-res)/p[i].s); 38 ans=res+(p[i].s*pos); 39 } 40 } 41 cout<<ans<<endl; 42 #ifndef ONLINE_JUDGE 43 cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl; 44 #endif 45 return 0; 46 }View Code