【枚举约数】HackerRank - Week of Code 26 - Satisfactory Pairs
2017-01-18 02:18
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题意:给你一个正整数n,问你存在多少个正整数对a,b(a<b),满足条件:存在正整数x,y,使得ax+by=n。
就预处理出n以内所有数的约数,然后暴力枚举a,暴力枚举x,然后枚举n-ax的所有约数,判重,统计答案即可。
就预处理出n以内所有数的约数,然后暴力枚举a,暴力枚举x,然后枚举n-ax的所有约数,判重,统计答案即可。
#include<cstdio> #include<vector> #include<algorithm> using namespace std; typedef vector<int>::iterator ITER; vector<int>divisors[300010]; int n,used[300010],ans; bool cmp(const int &a,const int &b) { return a>b; } int main() { scanf("%d",&n); for(int i=1;i<=n;++i) { for(int j=1;j*j<=i;++j) if(i%j==0) { divisors[i].push_back(j); if(j*j!=i) divisors[i].push_back(i/j); } sort(divisors[i].begin(),divisors[i].end(),cmp); } for(int a=1;a<n;++a)//枚举a for(int x=1;x*a<n;++x)//枚举x { int yb=n-x*a; for(ITER it=divisors[yb].begin();it!=divisors[yb].end();++it)//*it 就是b { if((*it)<=a)//因为b>a break; if(used[*it]!=a)//防止重复统计 { ++ans; used[*it]=a; } } } printf("%d\n",ans); return 0; }
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