Hartals

Hartals

时间: 1ms        内存:64M

描述:

Political parties in Bangladesh show their muscle by calling for regular hartals (strikes), which cause considerable economic damage. For our purposes, each party may be characterized by a positive integer h called the hartal parameter that denotes the average number of days between two successive strikes called by the given party. Consider three political parties. Assume h1 = 3, h2 = 4, and h3 = 8, where hi is the hartal parameter for party i. We can simulate the behavior of these three parties for N = 14 days. We always start the simulation on a Sunday. There are no hartals on either Fridays or Saturdays.

1 2 3 4 5 6 7 8 9 10 11 12 13 14
Days Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa
Party 1 x x x x
Party 2 x x x
Party 3 x
Hartals 1 2 3 4 5

There will be exactly five hartals (on days 3, 4, 8, 9, and 12) over the 14 days. There is no hartal on day 6 since it falls on Friday. Hence we lose five working days in two weeks. Given the hartal parameters for several political parties and the value of N, determine the number of working days lost in those N days.

输入:

The first line of the input consists of a single integer T giving the number of test cases to follow. The first line of each test case contains an integer N ( 7=<N<=3, 650), giving the number of days over which the simulation must be run. The next line contains another integer P ( 1P100) representing the number of political parties. The ith of the next P lines contains a positive integer hi (which will never be a multiple of 7) giving the hartal parameter for party i ( 1iP).

输出:

For each test case, output the number of working days lost on a separate line.

示例输入:

2
14
3
3
4
8
100
4
12
15
25
40

示例输出:

5
15

提示:

参考答案(内存最优[760]):

#include<stdio.h>
int a[3700];
int main()
{
	int t,n,num,x,i,count;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&num);
		for(i=0;i<=num;i++)
			a[i]=0;
		scanf("%d",&n);
		while(n--)
		{
			scanf("%d",&x);
			for(i=1;i*x<=num;i++)
				a[i*x]=1;
		}
		count=0;
		for(i=1;i<=num;i++)
		{
			if(i%7==6)
			{
				i++;
				continue;
			}
			else 
			{
				if(a[i])count++;
			}
		}
		printf("%d\n",count);
	}
	return 0;
}

参考答案(时间最优[0]):

#include<stdio.h>
int a[3700];
int main()
{
	int t,n,num,x,i,count;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&num);
		for(i=0;i<=num;i++)
			a[i]=0;
		scanf("%d",&n);
		while(n--)
		{
			scanf("%d",&x);
			for(i=1;i*x<=num;i++)
				a[i*x]=1;
		}
		count=0;
		for(i=1;i<=num;i++)
		{
			if(i%7==6)
			{
				i++;
				continue;
			}
			else 
			{
				if(a[i])count++;
			}
		}
		printf("%d\n",count);
	}
	return 0;
}

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