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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:
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