标签:2014acm辽宁省赛 acm algorithm 算法
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; }
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标签:2014acm辽宁省赛 acm algorithm 算法
原文地址:http://blog.csdn.net/u012997373/article/details/26629555