这套题最后一题不会，然后先放一下，最后一题应该是大数据结构题
A:求连续最长严格递增的的串，O（n）简单dp
#include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <iostream> #include <algorithm> #include <queue> #include <vector> using namespace std; typedef long long LL; const int N = 1e5+5; const int INF = 0x3f3f3f3f; int n,dp[N],a[N],cnt,mx; int main(){ scanf("%d",&n);a[0]=INF; for(int i=1;i<=n;++i){ scanf("%d",&a[i]); dp[i]=1; if(a[i]>a[i-1])dp[i]=dp[i-1]+1; mx=max(mx,dp[i]); } printf("%d ",mx); return 0; }
B:水题，map乱搞
#include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <iostream> #include <algorithm> #include <map> #include <queue> #include <vector> using namespace std; typedef long long LL; const int N = 1e5+5; const int INF = 2e9; map<int,int>mp; int a[40],cnt; int main(){ for(int i=0;;++i){ if((1ll<<i)>INF)break; a[++cnt]=(1<<i); } LL ret=0; int x,n;scanf("%d",&n); for(int i=1;i<=n;++i){ scanf("%d",&x); for(int j=cnt;j>0;--j){ if(a[j]<=x)break; int tmp=a[j]-x; if(mp.find(tmp)!=mp.end()) ret+=mp[tmp]; } if(mp.find(x)==mp.end())mp[x]=0; ++mp[x]; } printf("%I64d ",ret); return 0; }
C:一个典型的二分题，judge如何判断全被覆盖？只要用一下离线求和数组非0就好
#include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <iostream> #include <algorithm> #include <map> #include <queue> #include <vector> using namespace std; typedef long long LL; const int N = 1e5+5; const int INF = 2e9; LL a[N],b[N]; int n,m,c[N]; bool judge(LL r){ memset(c,0,sizeof(c)); for(int i=1;i<=m;++i){ int x=lower_bound(a+1,a+1+n,b[i]-r)-a; int y=upper_bound(a+1,a+1+n,b[i]+r)-a; ++c[x];--c[y]; } for(int i=1;i<=n;++i){ c[i]+=c[i-1]; if(!c[i])return false; } return true; } int main(){ scanf("%d%d",&n,&m); for(int i=1;i<=n;++i) scanf("%I64d",&a[i]); sort(a+1,a+1+n); n=unique(a+1,a+1+n)-a-1; for(int i=1;i<=m;++i) scanf("%I64d",&b[i]); sort(b+1,b+1+m); m=unique(b+1,b+1+m)-b-1; LL l=0,r=INF; while(l<r){ LL mid=(l+r)>>1; if(judge(mid))r=mid; else l=mid+1; } printf("%I64d ",(l+r)>>1); return 0; }
D：一个简单的分类讨论，因为最多走k，以k为周期即可
#include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <iostream> #include <algorithm> #include <map> #include <queue> #include <vector> using namespace std; typedef long long LL; const int N = 1e5+5; const int INF = 2e9; LL d,k,a,b,t; int main(){ scanf("%I64d%I64d%I64d%I64d%I64d",&d,&k,&a,&b,&t); if(d<=k){ printf("%I64d ",d*a); return 0; } LL ret=k*a;d-=k; if(d<=k){ ret+=min(d*a+t,d*b); printf("%I64d ",ret); return 0; } LL t1=k*a+t,t2=k*b; if(t2<=t1){ printf("%I64d ",ret+d*b); return 0; } else { ret+=d/k*t1; d-=d/k*k; if(d==0){printf("%I64d ",ret);return 0;} t1=t+d*a,t2=d*b; ret+=min(t1,t2); printf("%I64d ",ret); } return 0; }
E:求从每个点出发路径长度为k的边权和以及边权最小值，刚开始还以为是快速幂，结果发现发现这条路唯一确定，直接倍增即可
#include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <iostream> #include <algorithm> #include <map> #include <queue> #include <vector> using namespace std; typedef long long LL; const int N = 1e5+5; const int INF = 2e9; struct Node{ int v,mn; LL sum; }f[N][40]; LL retsum[N],k; int n,retmin[N],cur[N]; void solve(){ for(int j=0;(1ll<<j)<=k;++j)if(k&(1ll<<j)){ for(int i=0;i<n;++i){ retsum[i]+=f[cur[i]][j].sum; if(retmin[i]==-1)retmin[i]=f[cur[i]][j].mn; else retmin[i]=min(retmin[i],f[cur[i]][j].mn); cur[i]=f[cur[i]][j].v; } } for(int i=0;i<n;++i) printf("%I64d %d ",retsum[i],retmin[i]); } int main(){ scanf("%d%I64d",&n,&k); for(int i=0;i<n;++i)scanf("%d",&f[i][0].v),cur[i]=i,retmin[i]=-1; for(int i=0;i<n;++i)scanf("%d",&f[i][0].mn),f[i][0].sum=f[i][0].mn; for(int j=1;(1ll<<j)<=k;++j){ for(int i=0;i<n;++i){ f[i][j].v=f[f[i][j-1].v][j-1].v; f[i][j].sum=f[i][j-1].sum+f[f[i][j-1].v][j-1].sum; f[i][j].mn=min(f[i][j-1].mn,f[f[i][j-1].v][j-1].mn); } } solve(); return 0; }
F：不会，看了看别人的代码，并不能看懂，还是太弱