Educational Codeforces Round 96
A B C D E
英语阅读题。
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; inline int read(){ int f=1,ans=0;char c=getchar(); while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();} return f*ans; } int T,N; int main(){ T=read(); while(T--){ N=read();bool ff=0; for(int i=0;i*7<=N;i++){ for(int j=0;i*7+j*5<=N;j++){ int res=N-i*7-j*5; if((res%3)==0){ printf("%d %d %d\n",res/3,j,i); ff=1;break; } }if(ff) break; } if(!ff) printf("-1\n"); }return 0; }
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #define int long long using namespace std; inline int read(){ int f=1,ans=0;char c=getchar(); while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();} return f*ans; } const int MAXN=2e5+11; int T,N,A[MAXN],K; signed main(){ T=read(); while(T--){ N=read(),K=read();for(int i=1;i<=N;i++) A[i]=read(); sort(A+1,A+N+1);int ans=0; for(int i=N-K;i<=N;i++) ans+=A[i]; printf("%lld\n",ans); } }
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #include<queue> #define pii pair<int,int> #define fi first #define se second using namespace std; inline int read(){ int f=1,ans=0;char c=getchar(); while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();} return f*ans; } const int MAXN=2e5+11; pii Ans[MAXN]; int T,N,tot; priority_queue<int> que; int main(){ T=read(); while(T--){ tot=0;while(!que.empty()) que.pop(); N=read();for(int i=1;i<=N;i++) que.push(i); while(que.size()!=1){ int a=que.top();que.pop(); int b=que.top();que.pop(); Ans[++tot].fi=a,Ans[tot].se=b; int x=(a+b+1)/2; que.push(x); } printf("%d\n",que.top()); for(int i=1;i<=tot;i++) printf("%d %d\n",Ans[i].fi,Ans[i].se); } return 0; }
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; inline int read(){ int f=1,ans=0;char c=getchar(); while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();} return f*ans; } const int MAXN=2e5+11; int T,N,A[MAXN],F[MAXN]; char str[MAXN]; int main(){ T=read(); while(T--){ N=read();scanf("%s",str+1);F[0]=0; for(int i=1;i<=N;i++) A[i]=str[i]-'0'; int l=1; for(int i=2;i<=N;i++){ if(A[i]!=A[i-1]){ F[++F[0]]=i-l; l=i; } } F[++F[0]]=N-l+1;l=1;int ans=0; for(int i=1;i<=F[0];i++){ if(i>l) l=i; while(l<=F[0]&&F[l]==1) l++; if(l==F[0]+1){int res=F[0]-i+1;ans+=(res+1)/2;break;} F[l]--;ans++; }printf("%d\n",ans); }return 0; }/* 1 6 111010 */
#include<iostream> #include<cstring> #include<cstdio> #include<cstring> #include<vector> #include<queue> #include<algorithm> #include<climits> #include<bitset> #define pii pair<int,int> #define pb push_back #define mp make_pair #define fi first #define se second #define int long long using namespace std; inline int read(){ int f=1,ans=0;char c=getchar(); while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();} return f*ans; } const int MAXN=2e5+11; int N;vector<int> vec[31];char S[MAXN],T[MAXN]; int A[MAXN],ps[31]; struct BIT{ int val[MAXN]; int lowbit(int x){return x&-x;} void Add(int x){for(;x<=N;x+=lowbit(x)){val[x]++;}return;} int Query(int x){int res=0;for(;x;x-=lowbit(x)) res+=val[x];return res;} }bit; signed main(){ N=read();scanf("%s",S+1); for(int i=1;i<=N;i++) T[i]=S[N-i+1]; for(int i=1;i<=N;i++) vec[S[i]-'a'].pb(i); for(int i=1;i<=N;i++) A[i]=vec[T[i]-'a'][ps[T[i]-'a']++]; int res=0;for(int i=N;i>=1;i--){ res+=bit.Query(A[i]),bit.Add(A[i]); } printf("%lld\n",res);return 0; }
F
为啥 $n\leq 2000$ 啊，自闭了。
考虑手枪有剩余的情况什么时候会出现，一定时在 $l$ 的时刻换。
所以我们设 $f_i$ 表示第 $l_i$ 时必须手枪内至少有 $f_i$ 的子弹，直接转移即可。
#include<iostream> #include<cstring> #include<cstdio> #include<cstring> #include<vector> #include<queue> #include<algorithm> #include<climits> #include<bitset> #define int long long #define pii pair<int,int> #define pb push_back #define mp make_pair #define fi first #define se second using namespace std; inline int read(){ int f=1,ans=0;char c=getchar(); while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();} return f*ans; } const int MAXN=2e3+11; int f[MAXN],N,Ans,K,L[MAXN],R[MAXN],W[MAXN]; signed main(){ //freopen("F.in","r",stdin); N=read(),K=read();for(int i=1;i<=N;i++) L[i]=read(),R[i]=read(),W[i]=read(),Ans+=W[i]; for(int i=N;i>=1;i--){ int res=W[i];if(R[i]==L[i+1]) res+=f[i+1]; f[i]=max(0ll,res-K*(R[i]-L[i])); if(f[i]>K){printf("-1\n");return 0;} } int res=K;for(int i=1;i<=N;i++){ if(res<f[i]) Ans+=res,res=K; res=(res-W[i]%K+K)%K; }printf("%lld\n",Ans);return 0; }
G
将位独立后直接状压，每次计算后面的贡献即可。
#include<iostream> #include<cstring> #include<cstdio> #include<cstring> #include<vector> #include<queue> #include<algorithm> #include<climits> #include<bitset> #define pii pair<int,int> #define pb push_back #define mp make_pair #define fi first #define se second using namespace std; inline int read(){ int f=1,ans=0;char c=getchar(); while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();} return f*ans; } const int MAXN=19; int N,M,f[1<<MAXN],pre[1<<MAXN],W[MAXN],G[1<<MAXN],vis[MAXN],Sum[1<<MAXN]; vector<int> vec[MAXN]; int Ans[MAXN],g[1<<MAXN]; void print(int Sta){ if(!Sta) return; print(pre[Sta]);int now=Sta^pre[Sta]; ++Ans[0];for(int i=1;i<=N;i++) if(now&(1<<(i-1))) Ans[i]=Ans[0]; return; } int main(){ //freopen("G.in","r",stdin); N=read(),M=read(); for(int i=1;i<=M;i++){ int u=read(),v=read(),w=read();vec[u].pb(v); W[u]+=w,W[v]-=w;g[1<<(u-1)]|=(1<<(v-1)); } for(int i=0;i<(1<<N);i++) for(int j=1;j<=N;j++) if(i&(1<<(j-1))) g[i]|=g[(1<<(j-1))]; for(int i=0;i<(1<<N);i++){ memset(vis,0,sizeof(vis));for(int j=1;j<=N;j++) if(i&(1<<(j-1))) vis[j]=1,Sum[i]+=W[j]; for(int j=1;j<=N;j++) if(vis[j]) for(auto v:vec[j]) {if(vis[v]) G[i]=1;} }memset(f,127/3,sizeof(f));f[0]=0; for(int i=1;i<(1<<N);i++){ for(int S=i;S;S=(S-1)&i){ int Now=S^i,New=S; if(G[New]||((g[i]& (i^((1<<N)-1)) ))) continue; int cur=f[Now]+Sum[((1<<N)-1)^Now]; if(f[i]>cur) pre[i]=Now,f[i]=cur; } }print((1<<N)-1); for(int i=1;i<=N;i++) printf("%d ",Ans[i]);printf("\n"); return 0; }
