Repeat Number

Repeat Number

时间限制: 1 Sec  内存限制: 128 MB
提交: 28  解决: 9
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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

先求出所有 11 22...到1111111所有的repeat number取值，然后就好办了。

AC代码：

#include<iostream>using namespace std;int a[60],t;void init(){    int k;    t=0; k=1;    for(int i=0;i<5;i++){        k=k*10+1;        for(int j=1;j<=9;j++)            a[t++]=k*j;    }    a[t++]=1111111;}int main(){    init();    int x,y;    while(cin>>x>>y){        int i,p,q;        for(i=0;i<t;i++)            if(a[i]>=2*x){                q=i;                break;            }        if(i==t){            cout<<0<<endl;            continue;        }        for(i=t-1;i>=0;i--)            if(a[i]<=2*y){                p=i;                break;            }        if(i==-1){            cout<<0<<endl;            continue;        }        int sum=0;        for(i=q;i<=p;i++){            if(a[i]&1){                if(a[i]/2-x<y-a[i]/2)                    sum+=a[i]/2-x+1;                else                    sum+=y-a[i]/2;            }            else{                if(a[i]/2-x<y-a[i]/2)                    sum+=a[i]/2-x+1;                else                    sum+=y-a[i]/2+1;            }        }        cout<<sum<<endl;    }    return 0;}