标签:iostream fine div force can memset == cst double
题意:
给你n对数,求一个数,可以让他整除每一对数的其中一个
思路:
枚举第一对数的质因数,然后暴力
代码:
#include<iostream> #include<cstdio> #include<algorithm> #include<cmath> #include<cstring> #include<string> #include<stack> #include<queue> #include<deque> #include<set> #include<vector> #include<map> #include<functional> #define fst first #define sc second #define pb push_back #define mem(a,b) memset(a,b,sizeof(a)) #define lson l,mid,root<<1 #define rson mid+1,r,root<<1|1 #define lc root<<1 #define rc root<<1|1 #define lowbit(x) ((x)&(-x)) using namespace std; typedef double db; typedef long double ldb; typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> PI; typedef pair<ll,ll> PLL; const db eps = 1e-7; const int mod = 1e9+7; const int maxn = 2e7+100; const int maxm = 2e6+100; const int inf = 0x3f3f3f3f; const db pi = acos(-1.0); inline int read(){ int num; char ch; while((ch=getchar())<‘0‘ || ch>‘9‘); num=ch-‘0‘; while((ch=getchar())>=‘0‘ && ch<=‘9‘){ num=num*10+ch-‘0‘; } return num; } int c = 0; int p[maxn]; void d(int x){ for(int i=2;1ll*i*i<=x;i++)if(x%i==0){ p[c++]=i; while(x%i==0)x/=i; } if(x>1)p[c++]=x; } ll gcd(ll a, ll b){ return b == 0 ? a : gcd(b, a % b); } PLL pa[150000 + 100]; bool cmp(PLL a, PLL b){ return max(a.fst, a.sc) < max(b.fst, b.sc); } int main() { int n; scanf("%d", &n); for(int i = 0; i < n; i++){ scanf("%I64d %I64d", &pa[i].fst, &pa[i].sc); } //sort(pa, pa+n, cmp); d(pa[0].fst); d(pa[0].sc); if(n==1){ printf("%I64d", pa[0].fst); return 0; } for(int i = 0; i < c; i++){ int flg = 1; for(int j = 0; j < n && flg; j++){ if(pa[j].fst%p[i]!=0 && pa[j].sc%p[i]!=0) flg = 0; } if(flg){ printf("%d", p[i]); return 0; } } printf("-1"); return 0; }
codeforces 1025B Weakened Common Divisor(质因数分解)
标签:iostream fine div force can memset == cst double
原文地址:https://www.cnblogs.com/wrjlinkkkkkk/p/9505108.html