首页 > 代码库 > 华中农业大学第五届程序设计大赛 （7/12）
华中农业大学第五届程序设计大赛 （7/12）
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;}
华中农业大学第五届程序设计大赛 （7/12）
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