Codeforces Round #376 (Div. 2)

A 模拟

 #include <cstdio>
 #include <cstring>
 int Ans;
 char Str[];
 inline int Abs(int x) {return x>?x:-x;}
 inline int Min(int x,int y) {return x>y?y:x;}
 int main()
 {
     char u='a'; Ans=;
     scanf("%s",Str+);
     for (int i=;i<=strlen(Str+);i++)
         Ans+=Min(Abs(Str[i]-u),-Abs(Str[i]-u)),u=Str[i];
     printf("%d\n",Ans);
     return ;
 }

A

B 贪心

 #include <cstdio>
 int n,a[];
 bool Flag;
 int main()
 {
     scanf("%d",&n); Flag=true;
     for (int i=;i<=n;i++) scanf("%d",&a[i]);
     for (int i=;i<n;i++)
     {
         if (a[i]&)
         {
             if (a[i+]==)
             {
                 Flag=false;
                 break;
             }
             a[i+]--;
         }
     }
     if (a[n]&) Flag=false;
     if (Flag) puts("YES"); else puts("NO");
     return ;
 }

B

C 并查集+暴力 没写路径压缩T了好久..

 #include <cstdio>
 #include <cstring>
 #include <vector>
 #define pb push_back
 using namespace std;
 const int Maxn=;
 int Father[Maxn],c[Maxn],Last[Maxn],l[Maxn],r[Maxn],n,m,k,tot,Cnt[Maxn],Ans,num[Maxn];
 vector<int> V[Maxn];
 int GetFather(int x)
 {
     if (x==Father[x]) return x;
     return Father[x]=GetFather(Father[x]);
 }
 inline void Merge(int x,int y) {Father[GetFather(x)]=GetFather(y);}
 inline int Max(int x,int y) {return x>y?x:y;}
 int main()
 {
     scanf("%d%d%d",&n,&m,&k); tot=;
     for (int i=;i<=n;i++) scanf("%d",&c[i]);
     for (int i=;i<=n;i++) Father[i]=i;
     for (int i=;i<=m;i++) scanf("%d%d",&l[i],&r[i]),Merge(l[i],r[i]);
     for (int i=;i<=n;i++) if (Father[i]==i) num[i]=++tot;
     for (int i=;i<=n;i++) V[num[GetFather(i)]].pb(c[i]);
     Ans=;
     for (int i=;i<=tot;i++)
     {
         int Ret=,Tmp=V[i].size();
         for (int j=;j<Tmp;j++)    Cnt[V[i][j]]=;
         for (int j=;j<Tmp;j++)
         {
             Cnt[V[i][j]]=Cnt[V[i][j]]+;
             Ret=Max(Ret,Cnt[V[i][j]]);
         }
         Ans+=Tmp-Ret;
     }
     printf("%d",Ans);
     return ;
 }

C

D 模拟+看了好久的题意 就是一个操作是每个数+1，是的最终的序列为字典序，比较的是序列，比较相邻两个得出可以在的区间求交即可.

 #include <iostream>
 #include <cstring>
 #include <algorithm>
 #include <cstdio>
 #include <vector>
 #define pb push_back
 using namespace std;
 const int Maxn=;
 int n,c,Len[Maxn],cnt,B[Maxn],x;
 vector<int> V[Maxn];
 inline int Min(int x,int y) {return x>y?y:x;}
 void Work(int x,int y)
 {
     int LEN=Min(Len[x],Len[y]);
     for (int i=;i<LEN;i++)
     {
         if (V[x][i]==V[y][i]) continue;
         if (V[x][i]<V[y][i])
         {
             B[]++,B[c-V[y][i]+]--;
             B[c-V[x][i]+]++,B[c]--;
         } else B[c-V[x][i]+]++,B[c-V[y][i]+]--;
         cnt++; return;
     }
     if (Len[x]>Len[y])
     {
         puts("-1");
         exit();
     }
 }
 int main()
 {
     scanf("%d%d",&n,&c);
     for (int i=;i<=n;i++)
     {
         scanf("%d",&Len[i]);
         for (int j=;j<=Len[i];j++) scanf("%d",&x),V[i].pb(x);
         if (i!=) Work(i-,i);
     }
     int Sum=;
     for (int i=;i<c;i++)
     {
         Sum+=B[i];
         if (Sum==cnt)
         {
             printf("%d\n",i);
             return ;
         }
     }
     puts("-1");
     return ;
 }

D

E 直接模拟..

 #include<bits/stdc++.h>
 using namespace std;
 int n,tot,i,a[];
 inline int Max(int x,int y) {return x>y?x:y;}
 int main()
 {
     scanf("%d",&n);
     for(i=;i<=n;i++) scanf("%d",&a[i]),a[i]+=a[i-];
     tot=a[n];
     for(i=n-;i>;i--) tot=Max(tot,a[i]-tot);
     printf("%d\n",tot);
 }

E

F 枚举前缀就可以了

 #include <cstdio>
 #define LL long long
 const LL Maxn=;
 LL n,x,vis[Maxn+],Sum[Maxn+],Ans;
 int main()
 {
     scanf("%I64d",&n);
     for (LL i=;i<=n;i++) scanf("%I64d",&x),vis[x]++;
     for (LL i=;i<=Maxn;i++) Sum[i]=Sum[i-]+vis[i];
     Ans=;
     for (LL i=;i<=Maxn;i++)
         if (vis[i]>)
         {
             LL Res=,t=Maxn/i;
             for (LL j=;j<=t;j++) Res+=(Sum[j*i-]-Sum[(j!=)?j*i-i-:j*i-i])*(j-)*i;
             Res+=(Sum[Maxn]-Sum[t*i-])*t*i;
             if (Res>Ans) Ans=Res;
         }
     printf("%I64d\n",Ans);
     return ;
 }

F