Valentine's Day Round 1002 Misaki's Kiss again
2015-03-15 11:13
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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5175
思路:暴力
若a^b=c,则a^c=b
设k满足gcd(n,m)=k且n^m=k
所以k必然是n的约数,又因n^k=m
所以枚举所有n的约数,对于每个约数得到一个m,判断是否满足。
小心暴数据。
思路:暴力
若a^b=c,则a^c=b
设k满足gcd(n,m)=k且n^m=k
所以k必然是n的约数,又因n^k=m
所以枚举所有n的约数,对于每个约数得到一个m,判断是否满足。
小心暴数据。
#include <iostream> #include <iomanip> #include <cstdio> #include <algorithm> #include <cmath> #include <queue> #include <vector> #include <stack> #include <string> #include <cstring> using namespace std; typedef long long ll; const int maxn=11111111; const int INF=0x7fffffff; const int mod=1e7+7; #define LSON l,m,rt<<1 #define RSON m+1,r,rt<<1|1 #define ESP 1e-7 ll nc[55555], arr[55555]; int depart(ll x) { int k=0; for(ll i=1;i*i<=x;i++) { if(!(x%i)) { nc[k++]=i; if(i*i!=x) nc[k++]=x/i; } } return k; } ll gcd(ll a, ll b) { if(b==0) return a; return gcd(b, a%b); } int main() { ll n; int n_case=0; while(scanf("%lld", &n)!=EOF) { int sum=0; printf("Case #%d:\n", ++n_case); int tt=depart(n); for(int i=0;i<tt;i++) { ll m=n^nc[i]; if(m>=1 && m<=n && gcd(n, m)==nc[i]) arr[sum++]=m; } sort(arr, arr+sum); printf("%d\n", sum); if(sum>0) { for(int i=0;i<sum-1;i++) printf("%lld ", arr[i]); printf("%lld\n", arr[sum-1]); } else printf("\n"); } return 0; }
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