Codeforces Round #538 (Div. 2)

目录

• A.Got Any Grapes?
• B.Yet Another Array Partitioning Task
• C.Trailing Loves (or L'oeufs?)
• D.Flood Fill(区间DP)
• E.Arithmetic Progression(交互 二分 随机化)
• F.Please, another Queries on Array?(线段树 欧拉函数)

比赛链接

貌似最近平均难度最低的一场div2了...
但我没有把握住机会TAT
D题没往DP想 写模拟自闭了40多分钟...才发现是个傻逼区间DP
再多二十分钟就能调出F的傻逼错误了...

A.Got Any Grapes?

puts的返回值原来是0。。刚开始写的return !puts("NO");在样例上RE了一次= =

#include <set>
#include <map>
#include <cstdio>
#include <cctype>
#include <vector>
#include <cstring>
#include <algorithm>
#define pc putchar
#define gc() getchar()
typedef long long LL;


inline int read()
{
    int now=0,f=1;register char c=gc();
    for(;!isdigit(c);c=='-'&&(f=-1),c=gc());
    for(;isdigit(c);now=now*10+c-48,c=gc());
    return now*f;
}

int main()
{
    int x=read(),y=read(),z=read(),a=read(),b=read(),c=read();
    if(a<x) return puts("NO"),0;
    a-=x;
    if(a+b<y) return puts("NO"),0;
    int tot=a+b+c-y-z;
    if(tot<0) return puts("NO"),0;
    return puts("YES"),0;

    return 0;
}

B.Yet Another Array Partitioning Task

因为可以每隔\(m\)个分一段，所以选出最大的\(m*k\)个数，每隔\(m\)个分一段就行了。

#include <set>
#include <map>
#include <cstdio>
#include <cctype>
#include <vector>
#include <cstring>
#include <algorithm>
#define pc putchar
#define gc() getchar()
typedef long long LL;
const int N=2e5+5;

struct Node
{
    int v,p;
    bool operator <(const Node &x)const
    {
        return v==x.v?p<x.p:v>x.v;
    }
}A[N];

inline int read()
{
    int now=0,f=1;register char c=gc();
    for(;!isdigit(c);c=='-'&&(f=-1),c=gc());
    for(;isdigit(c);now=now*10+c-48,c=gc());
    return now*f;
}

int main()
{
    static int Ans[N];
    static bool vis[N];
    int n=read(),m=read(),K=read();
    for(int i=1; i<=n; ++i) A[i]=(Node){read(),i};
    std::sort(A+1,A+1+n);
    LL sum=0;
    for(int i=1; i<=K*m; ++i) vis[A[i].p]=1, sum+=A[i].v;
    int cnt=0;
    for(int i=1,t=0; i<=n; ++i)
        if(vis[i] && ++t==m)
            t=0, Ans[++cnt]=i;
    printf("%I64d\n",sum);
    for(int i=1; i<cnt; ++i) printf("%d ",Ans[i]);

    return 0;
}

C.Trailing Loves (or L'oeufs?)

D.Flood Fill(区间DP)

E.Arithmetic Progression(交互 二分 随机化)

F.Please, another Queries on Array?(线段树 欧拉函数)

标签：now,int,Codeforces,long,read,gc,Div,include,538
来源： https://www.cnblogs.com/SovietPower/p/10361309.html