今天实在累了,还有的题晚点补。。。。
题目链接:http://acm.hzau.edu.cn/problemset.php?page=3
题目:acm.hzau.edu.cn/5th.pdf
A:Little Red Riding Hood
题意:给你n朵花,每朵花有个权值,然后每次取花最少要间隔k朵,求权值最大;
思路:简单dp;
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<cstdio> #include<cmath> #include<string> #include<queue> #include<algorithm> #include<stack> #include<cstring> #include<vector> #include<list> #include<set> #include<map> using namespace std; #define ll long long #define pi (4*atan(1.0)) #define eps 1e-4 #define bug(x) cout<<"bug"<<x<<endl; const int N=1e6+10,M=1e6+10,inf=2147483647; const ll INF=1e18+10,mod=2147493647; ///数组大小 int a[N]; ll dp[N]; int main() { int n,k; int T; scanf("%d",&T); while(T--) { memset(dp,0,sizeof(dp)); scanf("%d%d",&n,&k); for(int i=1;i<=n;i++) scanf("%d",&a[i]); for(int i=1;i<=n;i++) if(i<=k)dp[i]=max(dp[i-1],1LL*a[i]); else dp[i]=max(dp[i-1],dp[i-k-1]+a[i]); printf("%lld\n",dp[n]); } return 0; }
D:gcd
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<cstdio> #include<cmath> #include<string> #include<queue> #include<algorithm> #include<stack> #include<cstring> #include<vector> #include<list> #include<set> #include<map> using namespace std; #define ll long long #define pi (4*atan(1.0)) #define eps 1e-4 #define bug(x) cout<<"bug"<<x<<endl; const int N=1e6+10,M=1e6+10,inf=2147483647; const ll INF=1e18+10,mod=2147493647; ///数组大小 ll MOD; struct Matrix { ll a[2][2]; Matrix() { memset(a,0,sizeof(a)); } void init() { for(int i=0;i<2;i++) for(int j=0;j<2;j++) a[i][j]=(i==j); } Matrix operator + (const Matrix &B)const { Matrix C; for(int i=0;i<2;i++) for(int j=0;j<2;j++) C.a[i][j]=(a[i][j]+B.a[i][j])%MOD; return C; } Matrix operator * (const Matrix &B)const { Matrix C; for(int i=0;i<2;i++) for(int k=0;k<2;k++) for(int j=0;j<2;j++) C.a[i][j]=(C.a[i][j]+1LL*a[i][k]*B.a[k][j])%MOD; return C; } Matrix operator ^ (const ll &t)const { Matrix A=(*this),res; res.init(); ll p=t; while(p) { if(p&1)res=res*A; A=A*A; p>>=1; } return res; } }; int main() { Matrix base; base.a[0][0]=1;base.a[0][1]=1; base.a[1][0]=1;base.a[1][1]=0; int T; scanf("%d",&T); while(T--) { int n,m,p; scanf("%d%d%d",&n,&m,&p); int x=__gcd(n+2,m+2); MOD=p; if(x<=2) printf("%d\n",1%p); else { Matrix ans=base^(x-2); printf("%lld\n",(ans.a[0][0]+ans.a[0][1])%MOD); } } return 0; }
E:One Stroke
题意:给你一棵二叉树,点有点权,每次往左或者往右走,求最长走的路,并且点权和小于k;
思路:官方题解,尺取,我的写法,树上二分,
对于一条链,枚举每个点为终点,vector存该点到根节点的前缀和,二分一下即可;
详见代码;
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<cstdio> #include<cmath> #include<string> #include<queue> #include<algorithm> #include<stack> #include<cstring> #include<vector> #include<list> #include<set> #include<map> using namespace std; #define ll long long #define pi (4*atan(1.0)) #define eps 1e-4 #define bug(x) cout<<"bug"<<x<<endl; const int N=1e6+10,M=1e6+10,inf=2147483647; const ll INF=1e18+10,mod=2147493647; ///数组大小 int n,ans,k,a[N]; vector<int>v; void dfs(int x) { int s=0,t=v.size()-1; int e=v.size()-1,ansq=-1; while(s<=e) { int mid=(s+e)>>1; if(v[t]-v[mid]<=k) { ansq=mid; e=mid-1; } else s=mid+1; } if(v[t]<=k)ans=max(ans,t+1); else ans=max(ans,t-ansq); int z=v[v.size()-1]; if(x*2<=n) { v.push_back(z+a[x<<1]); dfs(x<<1); v.pop_back(); } if(x*2+1<=n) { v.push_back(z+a[x<<1|1]); dfs(x<<1|1); v.pop_back(); } } int main() { int T; scanf("%d",&T); while(T--) { ans=0; v.clear(); scanf("%d%d",&n,&k); for(int i=1;i<=n;i++) scanf("%d",&a[i]); v.push_back(a[1]); dfs(1); if(ans)printf("%d\n",ans); else printf("-1\n"); } return 0; }
G:Sequence Number
题意:找出最远的i<=j&&a[i]<=a[j]的长度;
思路:这是一道排序可以过的题,也可以rmq+二分 最快的写法可以用单调栈做到O(n)
我是求后面的最大值后缀,二分后缀;
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<cstdio> #include<cmath> #include<string> #include<queue> #include<algorithm> #include<stack> #include<cstring> #include<vector> #include<list> #include<set> #include<map> using namespace std; #define ll long long #define pi (4*atan(1.0)) #define eps 1e-4 #define bug(x) cout<<"bug"<<x<<endl; const int N=1e5+10,M=1e6+10,inf=2147483647; const ll INF=1e18+10,mod=2147493647; int a[N],nex[N]; int main() { int n; while(~scanf("%d",&n)) { memset(nex,0,sizeof(nex)); for(int i=1;i<=n;i++) scanf("%d",&a[i]); for(int j=n;j>=1;j--) nex[j]=max(a[j],nex[j+1]); int ans=0; for(int i=1;i<=n;i++) { int s=i,e=n,pos=-1; while(s<=e) { int mid=(s+e)>>1; if(nex[mid]>=a[i]) pos=mid,s=mid+1; else e=mid-1; } ans=max(ans,pos-i); } printf("%d\n",ans); } return 0; }
J:Color Circle
题意:对于一个点,找长度大于4,相同字母,并且回到原点;
思路:暴力搜索;
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<cstdio> #include<cmath> #include<string> #include<queue> #include<algorithm> #include<stack> #include<cstring> #include<vector> #include<list> #include<set> #include<map> using namespace std; #define ll long long #define pi (4*atan(1.0)) #define eps 1e-4 #define bug(x) cout<<"bug"<<x<<endl; const int N=1e2+10,M=1e6+10,inf=2147483647; const ll INF=1e18+10,mod=2147493647; ///数组大小 char a[N][N],vis[N][N]; int n,m,ans; int xx[4]={0,1,0,-1}; int yy[4]={1,0,-1,0}; int check(int x,int y) { if(x<=0||x>n||y<=0||y>m) return 0; return 1; } void dfs(int x,int y,int dep) { if(ans)return; for(int i=0;i<4;i++) { int xxx=x+xx[i]; int yyy=y+yy[i]; if(check(xxx,yyy)&&a[xxx][yyy]==a[x][y]) { if(vis[xxx][yyy]&&dep-vis[xxx][yyy]+1>=4) { ans=1; } else if(!vis[xxx][yyy]) { vis[xxx][yyy]=dep; dfs(xxx,yyy,dep+1); vis[xxx][yyy]=0; } } } } int main() { while(~scanf("%d%d",&n,&m)) { memset(vis,0,sizeof(vis)); ans=0; for(int i=1;i<=n;i++) scanf("%s",a[i]+1); for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { dfs(i,j,1); if(ans)break; } if(ans)break; } if(ans)printf("Yes\n"); else printf("No\n"); } return 0; }
K:Deadline
题意:给你n个bug,每个bug最晚修复时间,一个程序猿需要一天修复一个bug,问需要多少个程序猿才能修复成功;
思路:开始sort一下,遍历过去超时;
后面想想发现>=n的数根本没有必要,一个程序员总是够的,所以遍历,标记小于n的就是;
这题只要想到思路还是很简单的,假设所有工程师每天都在修复bug,那么对天数记录bug的前缀和,O(n)得到答案max(pre[i]+i-1)/i)
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<cstdio> #include<cmath> #include<string> #include<queue> #include<algorithm> #include<stack> #include<cstring> #include<vector> #include<list> #include<set> #include<map> using namespace std; #define ll long long #define pi (4*atan(1.0)) #define eps 1e-4 #define bug(x) cout<<"bug"<<x<<endl; const int N=1e6+10,M=1e6+10,inf=2147483647; const ll INF=1e18+10,mod=2147493647; ///数组大小 int a[N],pre[N]; int main() { int n; while(~scanf("%d",&n)) { memset(pre,0,sizeof(pre)); for(int i=1;i<=n;i++) { scanf("%d",&a[i]); if(a[i]>=N-5)continue; pre[a[i]]++; } int ans=1; for(int i=1;i<=1000000;i++) { pre[i]=pre[i]+pre[i-1]; ans=max(ans,pre[i]/i+(pre[i]%i?1:0)); } printf("%d\n",ans); } return 0; }
L:Happiness
思路:找AB即可;
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<cstdio> #include<cmath> #include<string> #include<queue> #include<algorithm> #include<stack> #include<cstring> #include<vector> #include<list> #include<set> #include<map> using namespace std; #define ll long long #define pi (4*atan(1.0)) #define eps 1e-4 #define bug(x) cout<<"bug"<<x<<endl; const int N=3e3+10,M=1e6+10,inf=2147483647; const ll INF=1e18+10,mod=2147493647; char a[M]; int main() { int T,cas=1; scanf("%d",&T); while(T--) { scanf("%s",a+1); int n=strlen(a+1); int ans=0; for(int i=1;i<=n;i++) if(a[i]==‘A‘&&a[i+1]==‘B‘) ans++; printf("Case #%d:\n%d\n",cas++,ans); } return 0; }
时间: 2024-11-08 18:17:05