hdu 4627 The Unsolvable Problem【hdu2013多校3签到】
2013-07-30 20:59
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链接:
http://acm.hdu.edu.cn/showproblem.php?pid=4627
The Unsolvable Problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 243 Accepted Submission(s): 143
Problem Description
There are many unsolvable problem in the world.It could be about one or about zero.But this time it is about bigger number.
Given an integer n(2 <= n <= 109).We should find a pair of positive integer a, b so that a + b = n and [a, b] is as large as possible. [a, b] denote the least common multiplier of a, b.
Input
The first line contains integer T(1<= T<= 10000),denote the number of the test cases.
For each test cases,the first line contains an integer n.
Output
For each test cases,print the maximum [a,b] in a line.
Sample Input
3
2
3
4
Sample Output
1
2
3
Source
2013 Multi-University Training Contest 3
题意:
找最大的最小公倍数给你一个数 N , a+b = N ,找最大的 lcm(a,b)
思路:
奇数的随便就看出来了,取一半就好了偶数的,写个暴力程序,打下表也可以随便看出规律,取一半了
#include<stdio.h> #include<math.h> #include<algorithm> using namespace std; __int64 gcd(__int64 a, __int64 b) { return b == 0 ? a : gcd(b,a%b); } __int64 lcm(__int64 a, __int64 b){ return a/gcd(a,b)*b; } int main() { int T; __int64 n; scanf("%d", &T); while(T--) { scanf("%I64d", &n); __int64 a,b; __int64 ans = 0,tmp1,tmp2; if(n&1) ans = lcm(n/2,n/2+1); else { if(n == 2) ans = 1; else { __int64 c = n/2-1; tmp1 = lcm(c,n-c); tmp2 = lcm(c-1,(n-c+1)); ans = max(tmp1,tmp2); } } printf("%I64d\n", ans); } return 0; }
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