sgu491 Game for Little Johnny
2012-08-29 00:25
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题目链接:http://acm.sgu.ru/problem.php?contest=0&problem=491
题目意思: 对于 a*x+b*y=N的一个方程,给出N的值(1<=N<=100 000),求使得方程存在正解的正整数对(a, b)的个数(,其中a<b)
解题思路: 用O(n * log(n))来记录1~100 000的所有的约数,然后对a 从1到 N/2上遍历求符合条件的b的个数
//用时 937ms
题目意思: 对于 a*x+b*y=N的一个方程,给出N的值(1<=N<=100 000),求使得方程存在正解的正整数对(a, b)的个数(,其中a<b)
解题思路: 用O(n * log(n))来记录1~100 000的所有的约数,然后对a 从1到 N/2上遍历求符合条件的b的个数
//用时 937ms
//============================================================================ // Name : sgu491.cpp // Author : ssslpk // Version : // Copyright : Your copyright notice // Description : Hello World in C++, Ansi-style //============================================================================ #include<cstdio> #include<iostream> #include<algorithm> #include<string> #include<vector> #include<set> #include<ctime> #define maxn 100001 using namespace std; typedef vector<int>::iterator it; typedef vector<int>::reverse_iterator rit; vector<int> submul[maxn]; int temp[maxn]; int num[maxn*50]; void init(int n) { int cnt=0; for(int i=1;i*i<=n;i++) if(n%i==0) { if(i*i==n){temp[cnt++]=i;continue;} temp[cnt++]=i;temp[cnt++]=n/i; } sort(temp,temp+cnt); for(int i=0;i < cnt;i++) submul .push_back(temp[i]); } int main() { int n,i; for(i=1;i<maxn;i++) init(i); while(scanf("%d",&n)!=EOF) { int ans=0; for(int a=1;a<=n/2;a++) { int cnt=0; for(int j=a;j<n;j+=a) { int by=n-j; for(rit b=submul[by].rbegin();b!=submul[by].rend() && *b >a;b++) num[cnt++]=*b; } if(cnt==0)continue; sort(num,num+cnt); ans++; for(int i=1;i<cnt;i++) if(num[i]!=num[i-1])ans++; } printf("%d\n",ans); } return 0; }
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