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【Codeforces】Round #376 (Div. 2)

發(fā)布時間：2023/12/13 综合教程 21 生活家

生活随笔收集整理的這篇文章主要介紹了【Codeforces】Round #376 (Div. 2) 小編覺得挺不錯的,現(xiàn)在分享給大家,幫大家做個參考.

http://codeforces.com/contest/731

不發(fā)題面了,自己點鏈接

總結(jié)一下

考場上

原以為這次要加很多raiting。。。

但FST狗記鄧,只加了58rating

總結(jié)一下

ABC切得很快(保持)
B題WA了2發(fā)不應(yīng)該,沒有想清楚
F題寫了大暴力,但不優(yōu)美,雖然過了P,但就沒有想了,很嚴(yán)重問題,得意忘形,雖然知道FST,但有僥幸心里,真正考試就完了
以為拿到了分就不干事了,嚴(yán)重問題,真正考試一定要寫暴力拍
后來沒有干事,既沒有hack又沒有去寫DE不應(yīng)該

ANight at the Museum

Solution:Implementation,String

// <A.cpp> - Sun Oct 16 17:45:40 2016
// This file is made by YJinpeng，created by XuYike's black technology automatically.
// Copyright (C) 2016 ChangJun High School, Inc.
// I don't know what this program is.

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#define MOD 1000000007
#define INF 1e9
using namespace std;
typedef long long LL;
const int MAXN=110;
inline int gi() {
	register int w=0,q=0;register char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')q=1,ch=getchar();
	while(ch>='0'&&ch<='9')w=w*10+ch-'0',ch=getchar();
	return q?-w:w;
}
char s[MAXN];
int main()
{
	freopen("A.in","r",stdin);
	freopen("A.out","w",stdout);
	scanf("%s",s);int l=strlen(s);
    char ch='a';int ans=0;
    for(int i=0;i<l;i++)
        ans+=min((s[i]-ch+26)%26,(ch-s[i]+26)%26),ch=s[i];
    printf("%d",ans);
	return 0;
}

BCoupons and Discounts

題目大意:每天要買a[i]個東西,買的方式有兩種:兩天一天買一個 or 一天買兩個,問是否有合法方案

Solution:Greedy+遞推,考慮一天,奇數(shù)的話必須要用第一種方式,那就之后那天要買一個,幾個flag即可,一路推過去，<0不合法,偶數(shù)的話直接用第二種→ 貪心,不用麻煩后面那天。

反思:居然wa了兩發(fā),7~80分啊。。(notice 最后一次是否要之后填充,flag清空)

// <B.cpp> - Sun Oct 16 17:45:40 2016
// This file is made by YJinpeng，created by XuYike's black technology automatically.
// Copyright (C) 2016 ChangJun High School, Inc.
// I don't know what this program is.

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#define MOD 1000000007
#define INF 1e9
using namespace std;
typedef long long LL;
const int MAXN=100010;
const int MAXM=100010;
inline int gi() {
	register int w=0,q=0;register char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')q=1,ch=getchar();
	while(ch>='0'&&ch<='9')w=w*10+ch-'0',ch=getchar();
	return q?-w:w;
}
void pri(){printf("NO");exit(0);}
int main()
{
	freopen("B.in","r",stdin);
	freopen("B.out","w",stdout);
	int n=gi(),last=0;
    for(int i=1;i<=n;i++){
        int x=gi()-last;
        if(x<0)pri();
        if(x&1)last=1;else last=0;
    }if(last)pri();else printf("YES");
    return 0;
}

CSocks

有n只襪子,每只襪子有一個顏色,有一些約束(即某天的左右兩邊襪子顏色要相同),然后問你最少要改多少只?

顯然并查集,然后貪心,一個聯(lián)通塊內(nèi)的顏色要全相同,Greedy-找出顏色已經(jīng)相同的最多數(shù)目,答案加上其他要改的即可

// <C.cpp> - Sun Oct 16 17:45:40 2016
// This file is made by YJinpeng，created by XuYike's black technology automatically.
// Copyright (C) 2016 ChangJun High School, Inc.
// I don't know what this program is.

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#define MOD 1000000007
#define INF 1e9
using namespace std;
typedef long long LL;
const int MAXN=200010;
inline int gi() {
	register int w=0,q=0;register char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')q=1,ch=getchar();
	while(ch>='0'&&ch<='9')w=w*10+ch-'0',ch=getchar();
	return q?-w:w;
}
int f[MAXN],a[MAXN],c[MAXN];vector<int>b[MAXN];
int find(int x){return f[x]==x?x:f[x]=find(f[x]);}
int main()
{
	freopen("C.in","r",stdin);
	freopen("C.out","w",stdout);
	int n=gi(),m=gi(),k=gi(),ans=0;k++;
    for(int i=1;i<=n;i++)a[i]=gi(),f[i]=i;
    while(m--){
        int l=gi(),r=gi();
        if(find(l)!=find(r))f[f[l]]=f[r];
    }
    for(int i=1;i<=n;i++)
        b[find(i)].push_back(i);
    for(int i=1;i<=n;i++){
        if(f[i]!=i)continue;int now=0;
        for(int j=0,to=b[i].size();j<to;j++){
            c[a[b[i][j]]]++;
            if(c[a[b[i][j]]]>now)now=c[a[b[i][j]]];
        }
        ans+=b[i].size()-now;
        for(int j=0,to=b[i].size();j<to;j++)c[a[b[i][j]]]--;
    }printf("%d",ans);
	return 0;
}

D

挖坑~

E

挖坑~

FVideo Cards

給你一列數(shù)a[],選擇一個標(biāo)準(zhǔn),使得其他的數(shù)減小至他的倍數(shù),使得和最大?

CF給的標(biāo)簽:brute force(暴力)implementation(實現(xiàn))number theory(數(shù)論)-醉了
F是個送肉題。。。FST。。。。
直接看代碼吧
暴力也分優(yōu)美與垃圾
TLE on Test 23(P都過了,2×10^5都過了,結(jié)果FST,10^4沒過)-暴力加break

// <F.cpp> - Sun Oct 16 17:45:40 2016
// This file is made by YJinpeng，created by XuYike's black technology automatically.
// Copyright (C) 2016 ChangJun High School, Inc.
// I don't know what this program is.

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
using namespace std;
typedef long long LL;
const int MAXN=200010;
inline int gi() {
	register int w=0,q=0;register char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')q=1,ch=getchar();
	while(ch>='0'&&ch<='9')w=w*10+ch-'0',ch=getchar();
	return q?-w:w;
}
int a[MAXN];LL f[MAXN];
int main()
{
	int n=gi();LL ans=0;
    for(int i=1;i<=n;i++)a[i]=gi();
    sort(a+1,a+1+n);
    for(int i=1;i<=n;i++)f[i]=f[i-1]+a[i];
    for(int i=1;i<=n;i++){
        LL now=a[i];
        for(int j=n;j>i;j--){
            if(now+f[j]-f[i]<=ans)break;
            now+=a[j]/a[i]*a[i];
        }
        if(now>ans)ans=now;
    }printf("%I64d",ans);
	return 0;
}

AC-暴力+統(tǒng)計

// <F.cpp> - Sun Oct 16 17:45:40 2016
// This file is made by YJinpeng，created by XuYike's black technology automatically.
// Copyright (C) 2016 ChangJun High School, Inc.
// I don't know what this program is.


#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
using namespace std;
typedef long long LL;
const int MAXN=200001;
inline int gi() {
	register int w=0,q=0;register char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')q=1,ch=getchar();
	while(ch>='0'&&ch<='9')w=w*10+ch-'0',ch=getchar();
	return q?-w:w;
}
LL s[MAXN];int a[MAXN];
int main()
{
	freopen("F.in","r",stdin);
	freopen("F.out","w",stdout);
	int n=gi();LL ans=0;
    for(int i=1;i<=n;i++)a[gi()]++;
    for(int i=1;i<MAXN;i++)s[i]=s[i-1]+a[i];
    for(int i=1;i<MAXN;i++)
        if(a[i]){
            LL now=0;int t=MAXN/i;
            for(int j=1;j<t;j++)
                now+=1LL*(s[i*(j+1)-1]-s[i*j-1])*j*i;
            now+=1LL*(s[MAXN-1]-s[i*t-1])*i*t;
            if(now>ans)ans=now;
        }
    printf("%lld",ans);
	return 0;
}

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