A New Change Problem
Problem Description
Now given two kinds of coins A and B,which satisfy that GCD(A,B)=1.Here you can assume that there are enough coins for both kinds.Please calculate the maximal value that you cannot pay and the total number that you cannot pay.
Input
The input will consist of a series of pairs of integers A and B, separated by a space, one pair of integers per line.
Output
For each pair of input integers A and B you should output the the maximal value that you cannot pay and the total number that you cannot pay, and with one line of output for each line in input.
Sample Input
2 3
3 4
Sample Output
1 1
5 3
Author
lcy
题目类型：数论
算法分析：一个结论是：对于任意两个满足互质的a和b，其最大不能表示的数是a*b-(a+b)，不能表示的数的个数是(a-1)*(b-1)/2
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 |
#include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define lson rt << 1, l, m #define rson rt << 1 | 1, m + 1, r const int INF = 0x7FFFFFFF; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 1000000 + 66; int main() { // ifstream cin("aaa.txt"); long long a, b; while (cin >> a >> b) { cout << a * b - (a + b) << " " << (a - 1) * (b - 1) / 2 << endl; } return 0; } |
- « 上一篇：hdu1788
- hdu1846：下一篇 »