【训练】2017-11-8

最新推荐文章于 2024-11-14 14:37:58 发布
原创最新推荐文章于 2024-11-14 14:37:58 发布 · 427 阅读
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本文通过三个题目介绍了暴力算法的应用实例，包括日期判断、最大公约数计算及字符串操作等问题，并提供了源代码。
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早

T1

T1的话正解我也不知道是什么，但是呢，暴力能拿分，我手里页没有好的std，将就一下吧。
#include <cstdio>
#include <cstring>

char np[2022][13][32];

int ym[2][14]={
	{0,31,28,31,30,31,30,31,31,30,31,30,31,31},
	{0,31,29,31,30,31,30,31,31,30,31,30,31,31}
};

int isLeap(int yy){
	return (yy%400==0)||(yy%4==0&&yy%100!=0)?1:0;
}

void nextday(int &yy,int &mm,int &dd){
	if(mm==12&&dd==31){
		yy++;mm=1;dd=1;
		return;
	}
	dd++;
	if(dd>ym[isLeap(yy)][mm]){
		dd=1;
		mm++;
	}
}

bool hasTheSameDayOfNextMonth(int yy,int mm,int dd){
	return dd<=ym[isLeap(yy)][mm+1];
}

bool dfs(const int yy,const int mm,const int dd){
	if(np[yy][mm][dd]!=-1)return np[yy][mm][dd];
	if(
		yy>2012
	||	(yy==2012&&mm==12&&dd>22)
	)
		return np[yy][mm][dd]=true;
	if(yy==2012&&mm==12&&dd==22)
		return np[yy][mm][dd]=false;
	if(hasTheSameDayOfNextMonth(yy,mm,dd)){
		if(mm<12){
			if(!dfs(yy,mm+1,dd)){
				return np[yy][mm][dd]=true;
			}
		}else{
			if(!dfs(yy+1,1,dd)){
				return np[yy][mm][dd]=true;
			}
		}
	}
	int ty=yy,tm=mm,td=dd;
	nextday(ty,tm,td);
	if(!dfs(ty,tm,td))
		return np[yy][mm][dd]=true;
	return np[yy][mm][dd]=false;
}

int main(){
	freopen("calendar.in","r",stdin);
	freopen("calendar.out","w",stdout);
	int yy,mm,dd;
	memset(np,-1,sizeof(np));
	while(~scanf("%d%d%d",&yy,&mm,&dd)){
		printf("%s\n",dfs(yy,mm,dd)?"YES":"NO");
	}
	fclose(stdin); fclose(stdout);
	return 0;
}

T2

T2的话树状数组加暴力枚举，这个页没什么好说的，还是暴力嘛，我还是不会打。

我只有80P
#include<map>
#include<queue>
#include<cmath>
#include<cctype>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define qread(x) x=read()
#define mes(x,y) memset(x,y,sizeof(x))
#define mpy(x,y) memcpy(x,y,sizeof(x))
#define Maxn 500000
#define INF 2147483647  
inline int read(){
    char ch=getchar();
    int f=1,x=0;
    while(!(ch>='0'&&ch<='9')){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}
int n,m,x,y,z,sum[Maxn+1];
int main(){
    freopen("gcd.in","r",stdin);
    freopen("gcd.out","w",stdout);
    qread(n);qread(m);
    mes(sum,0);z=-INF;
    for(int i=1;i<=n;i++){
		qread(x);y=sqrt(x);
		for(int j=1;j<=y;j++){
			if(x%j==0){
				sum[j]++;
				sum[x/j]++;
			}
		}
		if(x==y*y)sum[y]--;
		z=std::max(z,x);
    }
    for(int i=z;i>=1;i--){
    	if(sum[i]>=m){
    		printf("%d\n",i*m);
    		return 0;
    	}
    }
}
/*
3 1
1 2 3
*/

 

std
#include <cstdio>

int n, k, x, m, cnt[500010];

int main(){
	freopen("gcd.in", "r", stdin); freopen("gcd.out", "w", stdout);
	for (scanf("%d%d", &n, &k); n--;) scanf("%d", &x), cnt[x]++, m = m < x ? x : m;
	for (int i = m, tmp; i; i--){
		tmp = 0;
		for (int j = i; j <= m; j += i) tmp += cnt[j];
		if (tmp >= k){
			printf("%I64d\n", (long long)i * k);
			fclose(stdin); fclose(stdout);
			return 0;
		}
	}
}

 

T3

T3其实还是暴力，我不会，就是把可以到达最前面且可以拉前来的就拉过来，贪心嘛。
#include<map>
#include<queue>
#include<cmath>
#include<cctype>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define qread(x) x=read()
#define mes(x,y) memset(x,y,sizeof(x))
#define mpy(x,y) memcpy(x,y,sizeof(x))
#define Maxn 100000
#define INF 2147483647  
inline int read(){
    char ch=getchar();
    int f=1,x=0;
    while(!(ch>='0'&&ch<='9')){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}
int n,m,tmp,len[2];
char s[Maxn+1];
int main(){
    freopen("pasuwado.in","r",stdin);
    freopen("pasuwado.out","w",stdout);
    qread(n);m=0;
    scanf("%s",s+1);len[0]=strlen(s+1);
    while(n>0&&m<len[0]){
    	len[1]=std::min(n+m+1,len[0]);
    	tmp=m+1;
    	for(int i=m+2;i<=len[1];i++)if(s[tmp]>s[i])tmp=i;
    	n-=tmp-m-1;
    	s[0]=s[tmp];
    	for(int i=tmp;i>=m+2;i--)s[i]=s[i-1];
    	s[m+1]=s[0];
    	m++;
    }
    s[len[0]+1]='\0';printf("%s\n",s+1);
}
/*
3
dcba
*/

 

std，std只有pas我很无奈啊。
var
	pk,tmp,now,tt,i,j,k:longint;
	n,m:int64;
	ans,s:ansistring;
	r:array['a'..'z']of longint;
	t:array[1..100000]of longint;
	w:array[1..100000,1..2]of longint;
	p:array[1..100000]of boolean;
	ch:char;
procedure ins(x:longint);
begin
	if x>n then exit;
	inc(t[x]);
	ins(x+x and (-x));
end;
function found(x:longint):longint;
begin
	if x=0 then exit(0);
	exit(t[x]+found(x-x and (-x)));
end;
procedure add(x:char;y:longint);
begin
	inc(tt);
	w[tt,1]:=y;
	w[tt,2]:=r[x];
	r[x]:=tt;
end;
begin
	assign(input,'pasuwado.in'); reset(input);
	assign(output,'pasuwado.out'); rewrite(output);
	readln(m);
	readln(s);
	n:=length(s); 
	for i:=n downto 1 do add(s[i],i);
	for now:=1 to n do begin
		for ch:='a' to 'z' do begin
			if r[ch]=0 then continue;
			tmp:=w[r[ch],1]-1-found(w[r[ch],1]);
			if tmp<=m then begin
				m:=m-tmp;
				ins(w[r[ch],1]);
				ans:=ans+s[w[r[ch],1]];
				r[ch]:=w[r[ch],2];
				break;
			end;
		end;
	end;
	writeln(ans);
	close(input); close(output);
end.

 

查看原文：http://hz2016.cn/blog/?p=121