A 模拟
1 #include <cstdio> 2 #include <cstring> 3 int Ans; 4 char Str[110]; 5 inline int Abs(int x) {return x>0?x:-x;} 6 inline int Min(int x,int y) {return x>y?y:x;} 7 int main() 8 { 9 char u='a'; Ans=0; 10 scanf("%s",Str+1); 11 for (int i=1;i<=strlen(Str+1);i++) 12 Ans+=Min(Abs(Str[i]-u),26-Abs(Str[i]-u)),u=Str[i]; 13 printf("%d\n",Ans); 14 return 0; 15 }
B 贪心
1 #include <cstdio> 2 int n,a[200100]; 3 bool Flag; 4 int main() 5 { 6 scanf("%d",&n); Flag=true; 7 for (int i=1;i<=n;i++) scanf("%d",&a[i]); 8 for (int i=1;i<n;i++) 9 { 10 if (a[i]&1) 11 { 12 if (a[i+1]==0) 13 { 14 Flag=false; 15 break; 16 } 17 a[i+1]--; 18 } 19 } 20 if (a[n]&1) Flag=false; 21 if (Flag) puts("YES"); else puts("NO"); 22 return 0; 23 }
C 并查集+暴力 没写路径压缩T了好久..
1 #include <cstdio> 2 #include <cstring> 3 #include <vector> 4 #define pb push_back 5 using namespace std; 6 const int Maxn=400100; 7 int Father[Maxn],c[Maxn],Last[Maxn],l[Maxn],r[Maxn],n,m,k,tot,Cnt[Maxn],Ans,num[Maxn]; 8 vector<int> V[Maxn]; 9 int GetFather(int x) 10 { 11 if (x==Father[x]) return x; 12 return Father[x]=GetFather(Father[x]); 13 } 14 inline void Merge(int x,int y) {Father[GetFather(x)]=GetFather(y);} 15 inline int Max(int x,int y) {return x>y?x:y;} 16 int main() 17 { 18 scanf("%d%d%d",&n,&m,&k); tot=0; 19 for (int i=1;i<=n;i++) scanf("%d",&c[i]); 20 for (int i=1;i<=n;i++) Father[i]=i; 21 for (int i=1;i<=m;i++) scanf("%d%d",&l[i],&r[i]),Merge(l[i],r[i]); 22 for (int i=1;i<=n;i++) if (Father[i]==i) num[i]=++tot; 23 for (int i=1;i<=n;i++) V[num[GetFather(i)]].pb(c[i]); 24 Ans=0; 25 for (int i=1;i<=tot;i++) 26 { 27 int Ret=0,Tmp=V[i].size(); 28 for (int j=0;j<Tmp;j++) Cnt[V[i][j]]=0; 29 for (int j=0;j<Tmp;j++) 30 { 31 Cnt[V[i][j]]=Cnt[V[i][j]]+1; 32 Ret=Max(Ret,Cnt[V[i][j]]); 33 } 34 Ans+=Tmp-Ret; 35 } 36 printf("%d",Ans); 37 return 0; 38 }
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.
1 #include <iostream> 2 #include <cstring> 3 #include <algorithm> 4 #include <cstdio> 5 #include <vector> 6 #define pb push_back 7 using namespace std; 8 const int Maxn=1000100; 9 int n,c,Len[Maxn],cnt,B[Maxn],x; 10 vector<int> V[Maxn]; 11 inline int Min(int x,int y) {return x>y?y:x;} 12 void Work(int x,int y) 13 { 14 int LEN=Min(Len[x],Len[y]); 15 for (int i=0;i<LEN;i++) 16 { 17 if (V[x][i]==V[y][i]) continue; 18 if (V[x][i]<V[y][i]) 19 { 20 B[0]++,B[c-V[y][i]+1]--; 21 B[c-V[x][i]+1]++,B[c]--; 22 } else B[c-V[x][i]+1]++,B[c-V[y][i]+1]--; 23 cnt++; return; 24 } 25 if (Len[x]>Len[y]) 26 { 27 puts("-1"); 28 exit(0); 29 } 30 } 31 int main() 32 { 33 scanf("%d%d",&n,&c); 34 for (int i=1;i<=n;i++) 35 { 36 scanf("%d",&Len[i]); 37 for (int j=1;j<=Len[i];j++) scanf("%d",&x),V[i].pb(x); 38 if (i!=1) Work(i-1,i); 39 } 40 int Sum=0; 41 for (int i=0;i<c;i++) 42 { 43 Sum+=B[i]; 44 if (Sum==cnt) 45 { 46 printf("%d\n",i); 47 return 0; 48 } 49 } 50 puts("-1"); 51 return 0; 52 }
E 直接模拟..
1 #include<bits/stdc++.h> 2 using namespace std; 3 int n,tot,i,a[200010]; 4 inline int Max(int x,int y) {return x>y?x:y;} 5 int main() 6 { 7 scanf("%d",&n); 8 for(i=1;i<=n;i++) scanf("%d",&a[i]),a[i]+=a[i-1]; 9 tot=a[n]; 10 for(i=n-1;i>1;i--) tot=Max(tot,a[i]-tot); 11 printf("%d\n",tot); 12 }
F 枚举前缀就可以了
1 #include <cstdio> 2 #define LL long long 3 const LL Maxn=200000; 4 LL n,x,vis[Maxn+100],Sum[Maxn+100],Ans; 5 int main() 6 { 7 scanf("%I64d",&n); 8 for (LL i=1;i<=n;i++) scanf("%I64d",&x),vis[x]++; 9 for (LL i=1;i<=Maxn;i++) Sum[i]=Sum[i-1]+vis[i]; 10 Ans=0; 11 for (LL i=1;i<=Maxn;i++) 12 if (vis[i]>0) 13 { 14 LL Res=0,t=Maxn/i; 15 for (LL j=1;j<=t;j++) Res+=(Sum[j*i-1]-Sum[(j!=1)?j*i-i-1:j*i-i])*(j-1)*i; 16 Res+=(Sum[Maxn]-Sum[t*i-1])*t*i; 17 if (Res>Ans) Ans=Res; 18 } 19 printf("%I64d\n",Ans); 20 return 0; 21 }
