noip模拟赛#45

T1：n<=1e6，求最小的区间包含(1,m)的所有数。

=>双指针扫一遍即可

#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
int read(){
	int x=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) x=x*10+c-'0',c=getchar();
	return x;
}
const int nmax=1e6+5;
const int maxn=2e3+5;
const int inf=0x7f7f7f7f;
int a[nmax],cnt[maxn];
int main(){
	freopen("Exhibit.in","r",stdin);freopen("Exhibit.out","w",stdout);
	int n=read(),m=read();rep(i,1,n) a[i]=read();
	int r=1,sm=0,ans=inf,res;
	rep(i,1,n){
		while(sm<m&&r<=n) {
			if(++cnt[a[r]]==1) ++sm;++r;
		}--r;
		//printf("(%d %d)\n",i,r);
		if(sm==m){
			if(ans>r-i+1) ans=r-i+1,res=i;
		}
		else break;
		if(r==n) break;
		if(!--cnt[a[i]]) --sm;++r;
	}
	printf("%d %d\n",res,res+ans-1);
	fclose(stdin);fclose(stdout);
	return 0;
}
/*
12 5
2 5 3 1 3 2 4 1 1 5 4 3
*/

T2：1000*1000的矩阵，求有多少个联通块。内存限制1M

=>懒得写暴力的了。。。正解似乎是并查集。我学的是ccz大爷的压位。就是压三位，其实多压几位也可以，char的范围是[-128,128]，然后就可以省下了一大堆空间。。然后ccz大爷写的bfs很好看。。。学了%%%

#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
char s[126][1001];
short q[4096][2];
#define push(x,y) ++qr,q[qr&4095][0]=x,q[qr&4095][1]=y,s[x>>3][y]-=(1<<(x&7))
void bfs(int x,int y){
	int ql=1,qr=1;q[1][0]=x;q[1][1]=y;
	while(ql<=qr){
		x=q[ql&4095][0],y=q[ql&4095][1];++ql;
		if(s[(x-1)>>3][y]&(1<<((x-1)&7))) push(x-1,y);
		if(s[(x+1)>>3][y]&(1<<((x+1)&7))) push(x+1,y);
		if(s[x>>3][y+1]&(1<<(x&7))) push(x,y+1);
		if(s[x>>3][y-1]&(1<<(x&7))) push(x,y-1); 
	}
}
int main(){
	freopen("Part.in","r",stdin);freopen("Part.out","w",stdout);
	int n,u,v,d,tmp,temp;scanf("%d",&n);
	rep(i,1,n) rep(j,1,n) {
		scanf("%d",&u),s[i>>3][j]|=(1-u)<<(i&7);
		//printf("(%d %d):%d\n",i,j,s[i>>3][j]);
	}
	int ans=0;
	rep(i,1,n) rep(j,1,n) if(s[i>>3][j]&(1<<(i&7))) bfs(i,j),++ans;
	printf("%d\n",ans);
	fclose(stdin);fclose(stdout);
	return 0;
}
/*
5
1 0 0 1 0
0 0 0 0 1
0 1 0 1 0
1 0 1 0 1
0 0 0 0 0
*/

T3：给一个数n，将n分解成若干个数的和。求最大的最小公倍数。答案<=1e25;

=>不会写。。。正解是利用一个性质就是2*3>2+3 f[2*3]<f[2+3]，所以每一个数必定只含有一个质因子。那么dp一下就可以了，然后再高精度。。

=>然后高精度很少写operator<的时候一直弄错。。。而且100以内的质数表自己手写写错了T_T。。。maya

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cctype>
using namespace std;
#define rep(i,s,t) for(ll i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define ll long long
const int a[]={0,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97};
struct node{
	int s[30],n;
	node(int _n=0){
		n=_n;clr(s,0);
	}
	node operator*(int k){
		node res(n);
		rep(i,1,n) res.s[i]=s[i]*k;
		rep(i,1,res.n) if(res.s[i]>=10) res.s[i+1]+=res.s[i]/10,res.s[i]%=10;
		if(res.s[n+1]) res.n++;
		for(;res.s[res.n];++res.n) if(res.s[res.n]>=10) {
			res.s[res.n+1]+=res.s[res.n]/10;
			res.s[res.n]%=10;
		}--res.n;
		return res;
	}
	node operator=(int x){
		for(n=0;x;x/=10) s[++n]=x%10;
		return *this;
	}
	node operator=(const node&o){
		n=o.n;
		rep(i,1,n) s[i]=o.s[i];
		return *this;
	}
	bool operator<(const node&o){
		if(n<o.n) return 1;
		if(n>o.n) return 0;
		dwn(i,n,1){
			if(s[i]<o.s[i]) return 1;
			if(s[i]>o.s[i]) return 0;
		}
		return 0;
	}
	void write(){
	    dwn(i,n,1) printf("%d",s[i]);printf("\n");
	}
};
node f[30][1200];
int main(){
	freopen("Lcm.in","r",stdin);freopen("Lcm.out","w",stdout);
    int n;scanf("%d",&n);
    rep(i,0,n) f[0][i]=1;
	rep(i,1,25){
		rep(j,a[i],n){
			f[i][j]=f[i-1][j];
			for(int k=a[i];k<=j;k=k*a[i]) if(f[i][j]<f[i-1][j-k]*k){
				 f[i][j]=f[i-1][j-k]*k;
			}
		}
	}
	node ans=0;
	rep(i,1,25) if(ans<f[i][n]) ans=f[i][n];
	ans.write();
	fclose(stdin);fclose(stdout);
	return 0;
}
/*
30
2 2 3 5 7 11
*/

 

转载于:https://www.cnblogs.com/fighting-to-the-end/p/5992690.html