Repeat Number

Repeat Number

题目描述

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].

输入

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.

输出

For each test output the number of couple of Repeat Number in one line.

样例输入

1 10
10 12

样例输出

5
2

提示

If a equals b, we can call a, b are Repeat Numbers too, and a is the Repeat Numbers for itself.

题意：a+b=c中c每一位数字都相同，为在x~y中有多少个这样的组合，其中数字可以重复，例如11+11==22；题解：找到x*2和y*2中c个区间，然后暴力求解；代码如下：

#include <stdio.h>#include <string.h>#include <algorithm>#include <math.h>#include <stdlib.h>using namespace std;int a[40];void f(){    int j,k=0;    for (int i = 11; i <= 10000000; i = i*10+1)        for (j = 1; j <= 9; j ++)            a[k ++] = i*j;}int main (){    int n,m,i,j,k;    f();    while (scanf("%d%d",&n,&m)!=EOF)    {        int x=-1,y=0;        for (i = 0; i < 54; i ++)        {            if (a[i] >= n*2&& x==-1 )                x = i;            if (a[i] <= m*2)            {                y = i;            }        }        int sum = 0;        for (i = x; i <= y; i ++)        {            for (j = n; j <= m; j ++)            {                if(a[i]-j <j)                    break;                if (a[i]-j >= n && a[i]-j <= m && a[i]-j >= j){                    sum ++;                }            }        }        printf("%d\n",sum);    }    return 0;}