Codeforces Round #376 (Div. 2)

A 模拟

#include <cstdio>
#include <cstring>
int Ans;
char Str[110];
inline int Abs(int x) {return x>0?x:-x;}
inline int Min(int x,int y) {return x>y?y:x;}
int main()
{
    char u='a'; Ans=0;
    scanf("%s",Str+1);
    for (int i=1;i<=strlen(Str+1);i++)
        Ans+=Min(Abs(Str[i]-u),26-Abs(Str[i]-u)),u=Str[i];
    printf("%d
",Ans);
    return 0;
}

B 贪心

#include <cstdio>
int n,a[200100];
bool Flag;
int main()
{
    scanf("%d",&n); Flag=true;
    for (int i=1;i<=n;i++) scanf("%d",&a[i]);
    for (int i=1;i<n;i++)
    {
        if (a[i]&1)
        {
            if (a[i+1]==0) 
            {
                Flag=false;
                break;
            }
            a[i+1]--;
        }
    }
    if (a[n]&1) Flag=false;
    if (Flag) puts("YES"); else puts("NO");
    return 0;    
}

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=400100;
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=0;
    for (int i=1;i<=n;i++) scanf("%d",&c[i]);
    for (int i=1;i<=n;i++) Father[i]=i;
    for (int i=1;i<=m;i++) scanf("%d%d",&l[i],&r[i]),Merge(l[i],r[i]);
    for (int i=1;i<=n;i++) if (Father[i]==i) num[i]=++tot;
    for (int i=1;i<=n;i++) V[num[GetFather(i)]].pb(c[i]);
    Ans=0;
    for (int i=1;i<=tot;i++)
    {
        int Ret=0,Tmp=V[i].size();
        for (int j=0;j<Tmp;j++)    Cnt[V[i][j]]=0;
        for (int j=0;j<Tmp;j++)
        {
            Cnt[V[i][j]]=Cnt[V[i][j]]+1;
            Ret=Max(Ret,Cnt[V[i][j]]);
        }
        Ans+=Tmp-Ret;
    }
    printf("%d",Ans);
    return 0;
}

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=1000100;
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=0;i<LEN;i++)
    {
        if (V[x][i]==V[y][i]) continue;
        if (V[x][i]<V[y][i])
        {
            B[0]++,B[c-V[y][i]+1]--;
            B[c-V[x][i]+1]++,B[c]--;
        } else B[c-V[x][i]+1]++,B[c-V[y][i]+1]--;
        cnt++; return;
    }
    if (Len[x]>Len[y])
    {
        puts("-1");
        exit(0);
    }
}
int main()
{
    scanf("%d%d",&n,&c);
    for (int i=1;i<=n;i++)
    {
        scanf("%d",&Len[i]);
        for (int j=1;j<=Len[i];j++) scanf("%d",&x),V[i].pb(x);
        if (i!=1) Work(i-1,i);        
    }
    int Sum=0;
    for (int i=0;i<c;i++)
    {
        Sum+=B[i];
        if (Sum==cnt) 
        {
            printf("%d
",i);
            return 0;
        }
    }
    puts("-1");
    return 0;
}

E 直接模拟..

#include<bits/stdc++.h>
using namespace std;
int n,tot,i,a[200010]; 
inline int Max(int x,int y) {return x>y?x:y;}
int main()
{
    scanf("%d",&n);
    for(i=1;i<=n;i++) scanf("%d",&a[i]),a[i]+=a[i-1];
    tot=a[n];
    for(i=n-1;i>1;i--) tot=Max(tot,a[i]-tot);
    printf("%d
",tot);
}

F 枚举前缀就可以了

#include <cstdio>
#define LL long long
const LL Maxn=200000;
LL n,x,vis[Maxn+100],Sum[Maxn+100],Ans;
int main()
{
    scanf("%I64d",&n);
    for (LL i=1;i<=n;i++) scanf("%I64d",&x),vis[x]++;
    for (LL i=1;i<=Maxn;i++) Sum[i]=Sum[i-1]+vis[i];
    Ans=0;
    for (LL i=1;i<=Maxn;i++) 
        if (vis[i]>0)
        {
            LL Res=0,t=Maxn/i;
            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;
            Res+=(Sum[Maxn]-Sum[t*i-1])*t*i;
            if (Res>Ans) Ans=Res;
        }
    printf("%I64d
",Ans);
    return 0;
}