best code-1001Poor Hanamichi
来源:互联网 发布:kali linux手机版 编辑:程序博客网 时间:2024/06/06 13:20
Poor Hanamichi
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 743 Accepted Submission(s): 275
Problem Description
Hanamichi is taking part in a programming contest, and he is assigned to solve a special problem as follow: Given a range [l, r] (including l and r), find out how many numbers in this range have the property: the sum of its odd digits is smaller than the sum of its even digits and the difference is 3.A integer X can be represented in decimal as:X=A n ×10 n +A n−1 ×10 n−1 +…+A 2 ×10 2 +A 1 ×10 1 +A 0 The odd dights areA 1 ,A 3 ,A 5 … andA 0 ,A 2 ,A 4 … are even digits.Hanamichi comes up with a solution, He notices that:10 2k+1 mod 11 = -1 (or 10),10 2k mod 11 = 1, So X mod 11 =(A n ×10 n +A n−1 ×10 n−1 +…+A 2 ×10 2 +A 1 ×10 1 +A 0 )mod11 =A n ×(−1) n +A n−1 ×(−1) n−1 +…+A 2 −A 1 +A 0 = sum_of_even_digits – sum_of_odd_digitsSo he claimed that the answer is the number of numbers X in the range which satisfy the function: X mod 11 = 3. He calculate the answer in this way : Answer = (r + 8) / 11 – (l – 1 + 8) / 11.Rukaw heard of Hanamichi’s solution from you and he proved there is something wrong with Hanamichi’s solution. So he decided to change the test data so that Hanamichi’s solution can not pass any single test. And he asks you to do that for him.
Input
You are given a integer T (1 ≤ T ≤ 100), which tells how many single tests the final test data has. And for the following T lines, each line contains two integers l and r, which are the original test data. (1 ≤ l ≤ r ≤10 18 )
Output
You are only allowed to change the value of r to a integer R which is not greater than the original r (and R ≥ l should be satisfied) and make Hanamichi’s solution fails this test data. If you can do that, output a single number each line, which is the smallest R you find. If not, just output -1 instead.
Sample Input
33 42 507 83
Sample Output
-1-180
#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#define mem(x) memset((x),0,sizeof((x)))using namespace std;int a[20];int len = 0;int main(){ //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); int T; cin>>T; long long l,r; while(T--) { //scanf("%lld%lld",&l,&r); cin>>l>>r; long long ans = -1; for(long long i = l; i <= r; i++) { long long temp = i; if(temp % 11 == 3) { int k = 0; mem(a); while(temp) { a[k++] = temp%10; temp /= 10; } int odd = 0; //ÆæÊý int even = 0; for(int j = 0; j < k; j += 2) { odd += a[j]; even += a[j+1]; } if(odd - even!= 3) { ans = i; break; } } } cout<< ans <<endl; } return 0;}
0 0
- best code-1001Poor Hanamichi
- hdu 4956 ( BestCoder #5 1001) Poor Hanamichi
- HDU 4956 Poor Hanamichi
- hdu4956 Poor Hanamichi
- Round #5 1001 Poor Hanamichi(原来这么简单啊)
- hdu 4956 Poor Hanamichi(bestcoder Round 6 1001)
- hdu 4956 Poor Hanamichi(暴力)
- hdu 4956 Poor Hanamichi(枚举)
- HDU—— 4956 Poor Hanamichi
- HDU 4956/BC 5A Poor Hanamichi
- hdu 4956 Poor Hanamichi BestCoder Round #5(数学题)
- [BestCoder Round #5] hdu 4956 Poor Hanamichi (数学题)
- HDU 4956 Poor Hanamichi(暴力)——BestCoder Round #5
- 【CUGBACM15级BC第五场 A】hdu 4956 Poor Hanamichi
- hdu 4956 Poor Hanamichi(BC.R#5) 读懂了题意就是水题/坑- -比赛中居然没有过
- poor
- hdoj4956Poor Hanamichi【暴力】
- Code Merging Best Known Method (C/C++)
- IOS学习笔记 -- 音频视频
- linux编程--shell
- MYSQL用法(十) 将某字段的值全部+5
- Android:Sqlitedatabase学习小结
- javascript window.close()无法关闭当前标签页面
- best code-1001Poor Hanamichi
- POJ 1159
- const char*, char const*, char*const的区别
- HTML5 04 表单
- android之 activity及activity栈详解
- 分支-03. 三天打鱼两天晒网(15)
- N-Queens II
- hdu 2082 母函数
- Html5 create thumbnail