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  • 2013暑期第一周周赛G题 错误总结

    Description

    Let's call a number k-good if it contains all digits not exceeding k (0, ..., k). You've got a number k and an array a containing n numbers. Find out how many k-good numbers are in a (count each number every time it occurs in array a).

    Input

    The first line contains integers n and k (1 ≤ n ≤ 100, 0 ≤ k ≤ 9). The i-th of the following n lines contains integer ai without leading zeroes (1 ≤ ai ≤ 109).

    Output

    Print a single integer — the number of k-good numbers in a.

    Sample Input

    Input
    10 6
    1234560
    1234560
    1234560
    1234560
    1234560
    1234560
    1234560
    1234560
    1234560
    1234560
    
    Output
    10
    
    Input
    2 1
    1
    10
    
    Output
    1
    代码
    #include<stdio.h>
    #include<string.h>
    int main()
    {
    	char s[105][105], k;
    	int n, a;
    	int i, j;
    	scanf("%d %c", &n, &k);
    	getchar(); //这个地方,要加getchar();
    	for (i = 0; i < n; i++)
    	{
    		gets(s[i]);//他的后面不需getchar();
    	}
    	a = k - '0';
    	//printf("a = %d
    ",a);
    	int len, sum = 0, x = 0, bleg = 0, t = 0;
    	for (i = 0; i < n; i++)
    	{
    		int f[11] = {0};//这个地方,标记数组必须要放到函数体内部,开始放在了外部,全局变量,每当第二次输入,若第二次长度小于第一次,则不会清零;
    		bleg = 0; //bleg需要在这里标记,不能放到外面
    		len = strlen(s[i]);		//printf("len s[%d] = %d
    ",i,len);
    		if (len < a + 1)
    		{
    			bleg = 0;
    		}
    		else
    		{
    			t = 0;
    			for (j = 0; j < len; j++)
    			{
    				int b = s[i][j] - '0';	//printf("b = %d
    ",b);
    
    				if (s[i][j] <= k && f[b] == 0)
    				{
    					t++;		//printf("	 t = %d
    ",t);
    					f[b] = 1;
    				}
    
    				if (t == a + 1)
    				{
    					bleg = 1;
    					break;
    				}
    			}
    		}
    		if (bleg == 1)
    		{
    			sum++;
    		}
    	}
    	printf("%d
    ", sum);
    	return 0;
    }
    
    


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  • 原文地址:https://www.cnblogs.com/tenlee/p/4420158.html
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