问题 C: Repeat Number
时间限制: 1 Sec 内存限制: 128 MB
提交: 23 解决: 7
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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
提示
If a equals b, we can call a, b are Repeat Numbers too, and a is the Repeat Numbers for itself.
这道题，我是暴力加二分过的，枚举c的可能值即可。然后求中间点与边界差的最小值即可
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
vector<int> G;
void init()
{
int k=1;
for(int i=0;i<6;i++)
{
k=k*10+1;
for(int j=1;j<=9;j++)
G.push_back(k*j);
}
}
int main()
{
int x,y;
init();
while(cin>>x>>y)
{
int p=lower_bound(G.begin(),G.end(),2*x)-G.begin();
int q=lower_bound(G.begin(),G.end(),2*y)-G.begin();
int ans=0;
//for(int i=0;i<10;i++)
// cout<<G[i]<<endl;
if(G[q]>2*y) q--;
// cout<<p<<" "<<q<<endl;
for(int i=p;i<=q;i++)
{
int mid = G[i]/2;
if (mid* 2 == G[i]) ans += mid-x < y-mid ? mid-x+1 : y-mid+1;
else ans+= mid-x+1 < y-mid ? mid-x+1 : y-mid;
}
cout<<ans<<endl;
}
return 0;
}