The 2018 ACM-ICPC Asia Qingdao Regional Contest, Online（四題簽到)

C題

用map超時，mmp map一直超時寫了兩個小時，單純的數組就行

思路：有n條指令，值爲r

存下執行第i條指令的時候的值r，若重複則輸出“No”

否則“Yes” 

//C
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>

using namespace std;
typedef long long llong;
const int MAXN = 10000 + 10;

typedef struct
{
	int p, v, k;
} node;

bool book[MAXN][260];

int main()
{
	int t, n, i, r, sgn;
	char str[7];
	int p, v, k;
	node a[MAXN];
	
	scanf("%d", &t);
	getchar();
	while(t--)
	{
		scanf("%d", &n);
		getchar();
		for(i = 1; i <= n; i++)
		{
			scanf("%s", str);
			getchar();
			if(str[1] == 'd')
			{
				scanf("%d", &k);
				getchar();
				p = 1;
				a[i] = (node){p, v, k};
				continue;
			}
			scanf("%d%d", &v, &k);
			getchar();
			if(str[1] == 'e')
			{
				p = 2;
			}
			else if(str[1] == 'n')
			{
				p = 3;
			}
			else if(str[1] == 'l')
			{
				p = 4;
			}
			else if(str[1] == 'g')
			{
				p = 5;
			}
			a[i] = (node){p, v, k};
		}
		
		memset(book, false, sizeof(book));
		i = 1; r = 0;
		book[i][r] = true;
		sgn = 1;
		while(i <= n)
		{
			p = a[i].p;
			v = a[i].v;
			k = a[i].k;
			if(p == 1)
			{
				r += k;
				r %= 256;
				i++;
			}
			else if(p == 2)
			{
				i = (r == v) ? k : i + 1;
			}
			else if(p == 3)
			{
				i = (r != v) ? k : i + 1;
			}
			else if(p == 4)
			{
				i = (r < v) ? k : i + 1;
			}
			else if(p == 5)
			{
				i = (r > v) ? k : i + 1;
			}
			if(book[i][r] == true)
			{
				sgn = 0;
				break;
			}
			book[i][r] = true;
		}
		if(sgn == 1)
		{
			printf("Yes\n");
		}
		else
		{
			printf("No\n");
		}
	}
    return 0;
}

 

K

 異或運算，要求找到數量最多的集合，集合中任意兩個數異或值皆小於這兩個數。

 

//K
#include <bits/stdc++.h> 
using namespace std;
int a[100];
bool cmp(int c,int d)
{
	return c>d;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int tp,nu;
		memset(a,0,sizeof(a));
		scanf("%d",&nu);
		while(nu--)
		{
					
		scanf("%d",&tp);
		int cnt=0;
		while(tp!=0)
		{
			tp>>=1;
			cnt++;
		}
		a[cnt]++;
		
	    }
	    sort(a,a+100,cmp);
	    printf("%d\n",a[0]);
	}
	return 0;
}