Definition: a+b = c, if all the digits of c are same ( c is more than ten)，then we call a and b are Repeat Number. My question is How many Repeat Numbers in [x,y].
Definition: a+b = c, if all the digits of c are same ( c is more than ten)，then we call a and b are Repeat Number. My question is How many Repeat Numbers in [x,y].
There are several test cases.
Each test cases contains two integers x, y(1<=x<=y<=1,000,000) described above.
Proceed to the end of file.
For each test output the number of couple of Repeat Number in one line.
1 10
10 12
5
2
If a equals b, we can call a, b are Repeat Numbers too, and a is the Repeat Numbers for itself.
上代码
#include<stdio.h> int a[123]={ 11,22,33,44,55,66,77,88,99, 111,222,333,444,555,666,777,888,999, 1111,2222,3333,4444,5555,6666,7777,8888,9999, 11111,22222,33333,44444,55555,66666,77777,88888,99999, 111111,222222,333333,444444,555555,666666,777777,888888,999999, 1111111,2222222,3333333,4444444,5555555,6666666,7777777,8888888,9999999 }; int lower_bound(int *array, int size, int key) { int first = 0, middle; int half, len; len = size; while(len > 0) { half = len >> 1; middle = first + half; if(array[middle] < key) { first = middle + 1; len = len-half-1; } else len = half; } return first; } int main() { int x,y,i,mid; while(~scanf("%d%d",&x,&y)) { int p=lower_bound(a,54,2*x); int q=lower_bound(a,54,2*y); int ans=0; if(a[q]>2*y) q--; for(i=p;i<=q;i++) { mid = a[i]/2; if (a[i]%2==0) { if(mid-x < y-mid) ans+=mid-x+1; else ans+=y-mid+1; } else { if(mid-x+1 < y-mid) ans+=mid-x+1; else ans+=y-mid; } } printf("%d\n",ans); } return 0; } /************************************************************** Problem: 2 User: hui Language: C Result: 正确 Time:0 ms Memory:964 kb ****************************************************************/
原文：http://www.cnblogs.com/suthui/p/3816278.html