标签:暴力搜索
1.题目描述:点击打开链接
2.解题思路:根据题目描述,可以解出来y的范围是1≤y≤2k。进而可以求出x=ky/(y-n)。注意x要大于0且是整数。
3.代码:
#define _CRT_SECURE_NO_WARNINGS #include<iostream> #include<algorithm> #include<string> #include<sstream> #include<set> #include<vector> #include<stack> #include<map> #include<queue> #include<deque> #include<cstdlib> #include<cstdio> #include<cstring> #include<cmath> #include<ctime> #include<functional> using namespace std; #define N 50000 int u[N], v[N]; int main() { //freopen("t.txt", "r", stdin); int n; while (~scanf("%d", &n)) { int cnt = 0; int len = 2 * n; for (int y = 1; y <= len; y++) { if (y - n > 0 && n*y % (y - n) == 0) { int x = n*y / (y - n); u[cnt] = x, v[cnt++] = y; } } printf("%d\n", cnt); for (int i = 0; i < cnt; i++) printf("1/%d = 1/%d + 1/%d\n", n, u[i], v[i]); } return 0; }
标签:暴力搜索
原文地址:http://blog.csdn.net/u014800748/article/details/44888889