紫书 例题7-3 UVA 10976

【题意】给你一个数k，求所有使得1/k = 1/x + 1/y成立的x≥y的整数对。

【解题方法】数论，枚举。枚举所有在区间（k+1，2k）上的y即可，当1/k - 1/y的结果分子为1即为一组解。

【AC代码】

////Created by just_sort 2016/1/3//Copyright (c) 2016 just_sort.All Rights Reserved//#include <ext/pb_ds/assoc_container.hpp>#include <ext/pb_ds/tree_policy.hpp>#include <ext/pb_ds/hash_policy.hpp>#include <set>#include <map>#include <queue>#include <stack>#include <cmath>#include <cstdio>#include <time.h>#include <cstdlib>#include <cstring>#include <sstream> //isstringstream#include <iostream>#include <algorithm>using namespace std;using namespace __gnu_pbds;typedef long long LL;typedef pair<int, LL> pp;#define REP1(i, a, b) for(int i = a; i < b; i++)#define REP2(i, a, b) for(int i = a; i <= b; i++)#define REP3(i, a, b) for(int i = a; i >= b; i--)#define CLR(a, b)     memset(a, b, sizeof(a))#define MP(x, y)      make_pair(x,y)const int maxn = 100;const int maxm = 2e5;const int maxs = 10;const int INF  = 1e9;const int UNF  = -1e9;const int mod  = 1e9+7;int gcd(int x, int y) {return y == 0 ? x : gcd(y, x % y);}//typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>order_set;//headint a[10010];int main(){    int k;    while(scanf("%d", &k) != EOF)    {        int cnt = 0;        REP2(y, 1, 2 * k){            if(y == k) continue;            if((abs(k * y) % abs(y - k) == 0) && ((k * y) / (y - k) >= y)){                a[++cnt] = y;            }        }        printf("%d\n", cnt);        REP2(i, 1, cnt){            printf("1/%d = 1/%d + 1/%d\n", k, (k * a[i]) / (a[i] - k), a[i]);        }    }    return 0;}