| ||||||||||
| Online Judge | Problem Set | Authors | Online Contests | User | ||||||
|---|---|---|---|---|---|---|---|---|---|---|
| Web Board Home Page F.A.Qs Statistical Charts | Current Contest Past Contests Scheduled Contests Award Contest | |||||||||
将大牛的修改了下,加了点注释,这样可能好理解点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;
}
Followed by:
Post your reply here: |
All Rights Reserved 2003-2013 Ying Fuchen,Xu Pengcheng,Xie Di
Any problem, Please Contact Administrator