ZOJ 3629 Treasure Hunt IV

题意：若n满足求和 （k从1-n）[n/k]  结果是偶数，则n称为xx数 

给定a，b 问区间内是xx数的有多少个 ([]是取整符号)

数论不大好，就直接找规律做了

 

#include<stdio.h>#include<set>#include<math.h>using namespace std;#define N 10000000#define F(x) 2*x*x-x//F(0)=0#define ll long long#define maxx 9223372036854775807ll sec(ll x){return 2*(sqrt((double)x));}ll Ans(ll x){if(x==-1)return 0;ll k=(x/4);k=sqrt((double)k);ll ans=F(k);ll z=k*4;if(x>k*k*4+z)ans+=z+1;else ans+=x-k*k*4+1;return ans;}int main(){ll a,b;while(~scanf("%lld%lld",&a,&b))printf("%lld\n",Ans(b)-Ans(a-1));return 0;}/*bool can(int x){//打表代码int ans=0,i;for(i=1;i<=x;i++)ans+=x/i;return ans&1;}int main(){ll i,j=4,k;for(i=0;i<N;i++)if(!can(i))printf("%d\n",i);return 0;}*/