The military training of HH (allowance)

1248:HH Military Training time limit: 1 Sec memory limit: MB submitted: 11 [Submit] [status] [discussion] Title Description

University, the most memorable thing is military training, white HH children's shoes are ruthless tanning, saying that during military training, the most feared is to be instructors tease, as the saying goes: Cherish life, away from instructors. Instructors always love to let students stand square, because a lot of classmates, will inevitably have the students in front of the students in the back, in order to stay away from the instructor, we all hope that their position instructor can not see themselves, now if the students stand in a coordinate point of the matrix, the students are standing in each group of integers, and the instructor standing in the coordinates Please tell me how many positions instructors can see directly at this time!

Input

The first line enters a T, which represents the T-group test data.

Next, enter two digital m,n per line, and for any one of the students coordinate position (x, y), there are 1<=x<=m<=10000,1<=y<=n<=10000.

Output

Output How many positions instructors can see.

Sample input

21 42 3

Sample output

45
The puzzle: wa ... No words, the wrong afternoon ... Or WA, put it on first;
Code:

#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include < Algorithm> #include <vector>using namespace std; #define MEM (x, y) memset (x,y,sizeof (×)) typedef long Long LL; Vector<int>p;void getp (int x) {for (int i=2;i*i<x;i++) {if (x%i==0) {p.push_back (i), while (x%i==0) x/=i;}} if (x>1) p.push_back (x);} int getn (int x) {int sum=0;/*for (int i=0;i<p.size (); i++) {cur=p[i];sum+=x/cur;for (int j=i+1;j<p.size (); j + +) {cur *=-p[j];sum+=x/cur;}} *///This method does not work if it is 2,3,4, then 2, 4 can not be selected .... for (int i=1;i< (1<<p.size)), ++i) {int num=0,mult=1;for (int k=0;k< (int) p.size (); k++) {if (i& (1< <k)) {mult*=p[k];num++;}} if (num&1) sum+=x/mult;//is missing the equals sign: else Sum-=x/mult;} printf ("%d\n", sum); return sum;} int main () {int t,m,n;scanf ("%d", &t), while (t--) {scanf ("%d%d", &m,&n); int ans=m;for (int i=2;i<=n;i++) { P.clear (); Getp (i); Ans+=m-getn (M);} printf ("%d\n", ans);} return 0;}

　　

HH's military training (allowance)