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将大牛的修改了下,加了点注释,这样可能好理解点

Posted by fbixiaozc135 at 2007-08-08 15:36:25 on Problem 1019
In Reply To:Re:分段,再一个一个区间把N对应的数字套住 Posted by:fbixiaozc135 at 2007-08-08 15:35:24
#include <string.h>
#include <stdio.h>

int main()
{
    int a[5]={0, 45, 9045, 1395495, 189414495};
    int b[5]={0, 9, 189, 2889, 38889};
    int c[5]={0, 10, 100, 1000, 10000};
    unsigned long i,j,k,n,t;
    scanf("%d",&t);
	while(t--)
    {
        scanf("%d",&n);
        for ( i = 1; i < 5; i++)
        if (n <= a[i]) break;                 //所在的行数为 i位数
        n -= a[i - 1];
        for (k = 0, j = 1; k < n; j++)
            k = b[i - 1] * j + i * j * (j + 1) / 2;  //等差数列求和公式,找到具体的区间位置    
        --j;           
        n = n - b[i - 1] * (j - 1) - i * j * (j - 1) / 2;     //在该数中寻找最终的位置 
        for(i = 1; i < 5; i++)
        if (n <= b[i]) break;                //通过位置来确定最终这个数是几位数
        if (i == 1)         //如果是1位数,直接输出
			printf("%d\n",n);
        else                              //求得在该数的一排中 最后更新了的序号n对应的数字
        {
          n -= b[i-1];   
	/*要找的数最终在的位置的真正的数比如11的位置是4,这里k就等于11 */
          k = (n-1) / i + c[i-1];
		 char str[10];
         sprintf(str,"%d",k); //将数字转换成字符串
         n -= (n - 1) / i * i; 
		 printf("%c\n",str[n - 1]);
        }
    }    
    return 0;
}

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