題目
2301: [HAOI2011]Problem b
Time Limit: 50 Sec Memory Limit: 256 MBDescription
對於給出的n個詢問,每次求有多少個數對(x,y),滿足a≤x≤b,c≤y≤d,且gcd(x,y) = k,gcd(x,y)函數爲x和y的最大公約數。
Input
第一行一個整數n,接下來n行每行五個整數,分別表示a、b、c、d、k
Output
共n行,每行一個整數表示滿足要求的數對(x,y)的個數
Sample Input
2
2 5 1 5 1
1 5 1 5 2
2 5 1 5 1
1 5 1 5 2
Sample Output
14
3
HINT
100%的數據滿足:1≤n≤50000,1≤a≤b≤50000,1≤c≤d≤50000,1≤k≤50000
題解
這道題目,我自己推得公式是sigma{miu[p]*(b/kp-(a-1)/kp)*(d/kp-(c-1)/kp)},這樣是O(nd)的複雜度。還是TLE了,我不明覺厲,果然還是太弱!然後看網上的題解真是ORZ啊!
代碼
/*Author:WNJXYK*/
#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define LL long long
#define Inf 2147483647
#define InfL 10000000000LL
inline void swap(int &x,int &y){int tmp=x;x=y;y=tmp;}
inline void swap(LL &x,LL &y){LL tmp=x;x=y;y=tmp;}
inline int remin(int a,int b){if (a<b) return a;return b;}
inline int remax(int a,int b){if (a>b) return a;return b;}
inline LL remin(LL a,LL b){if (a<b) return a;return b;}
inline LL remax(LL a,LL b){if (a>b) return a;return b;}
const int Maxn=50000;
int miu[Maxn+10];
inline void getMiu(){
for (int i=1;i<=Maxn;i++){
int target=i==1?1:0;
int delta=target-miu[i];
miu[i]=delta;
for (int j=i+i;j<=Maxn;j+=i) miu[j]+=delta;
}
for (int i=2;i<=Maxn;i++) miu[i]+=miu[i-1];
}
inline LL getAns(int n,int m){
if (n>m) swap(m,n);
LL last;
LL ret=0;
for (LL i=1;i<=n;i=last+1){
last=remin(n/(n/i),(m/(m/i)));
ret+=((LL)miu[last]-(LL)miu[i-1])*(m/i)*(n/i);
}
return ret;
}
int T;
int a,b,c,d,k;
LL Ans=0;
int main(){
getMiu();
scanf("%d",&T);
for (;T--;){
scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
a-=1;c-=1;
a/=k;b/=k;c/=k;d/=k;
Ans=getAns(b,d)+getAns(a,c)-getAns(a,d)-getAns(b,c);
printf("%lld\n",Ans);
}
return 0;
}