import sys
readline=sys.stdin.readline
from collections import Counter
class Prime:
def __init__(self,N):
assert N<=10**8
self.smallest_prime_factor=[None]*(N+1)
for i in range(2,N+1,2):
self.smallest_prime_factor[i]=2
n=int(N**.5)+1
for p in range(3,n,2):
if self.smallest_prime_factor[p]==None:
self.smallest_prime_factor[p]=p
for i in range(p**2,N+1,2*p):
if self.smallest_prime_factor[i]==None:
self.smallest_prime_factor[i]=p
for p in range(n,N+1):
if self.smallest_prime_factor[p]==None:
self.smallest_prime_factor[p]=p
self.primes=[p for p in range(N+1) if p==self.smallest_prime_factor[p]]
def Factorize(self,N):
assert N>=1
factors=defaultdict(int)
if N<=len(self.smallest_prime_factor)-1:
while N!=1:
factors[self.smallest_prime_factor[N]]+=1
N//=self.smallest_prime_factor[N]
else:
for p in self.primes:
while N%p==0:
N//=p
factors[p]+=1
if N<p*p:
if N!=1:
factors[N]+=1
break
if N<=len(self.smallest_prime_factor)-1:
while N!=1:
factors[self.smallest_prime_factor[N]]+=1
N//=self.smallest_prime_factor[N]
break
else:
if N!=1:
factors[N]+=1
return factors
def Divisors(self,N):
assert N>0
divisors=[1]
for p,e in self.Factorize(N).items():
pow_p=[1]
for _ in range(e):
pow_p.append(pow_p[-1]*p)
divisors=[i*j for i in divisors for j in pow_p]
return divisors
def Is_Prime(self,N):
return N==self.smallest_prime_factor[N]
def Totient(self,N):
for p in self.Factorize(N).keys():
N*=p-1
N//=p
return N
def Mebius(self,N):
fact=self.Factorize(N)
for e in fact.values():
if e>=2:
return 0
else:
if len(fact)%2==0:
return 1
else:
return -1
T=int(readline())
P=Prime(10**6)
for t in range(T):
N,M=map(int,readline().split())
cnt=[0]*(N+1)
for g in range(1,N+1):
cnt[g]=N//g*(N//g-1)//2
for p in P.primes:
if N<p:
break
for g in range(p,N+1,p):
cnt[g//p]-=cnt[g]
ans=0
for g in range(N,1,-1):
c=min(cnt[g],M)//(g-1)
ans+=c*g
M-=c*(g-1)
if M:
ans=-1
print(ans)
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
int TC, n;
ll m;
cin >> TC;
while (TC--){
scanf("%d%lld", &n, &m);
ll dp[n+1], res=0;
for (int d = n; d >= 2; d--){
dp[d] = ((n/d)*1ll*((n/d)-1))/2;
for (int k = 2; k <= n/d; k++) dp[d] -= dp[k*d];
}
for (ll d = n; d >= 2; d--){
ll t = min(m/(d-1), dp[d]/(d-1));
m-=(d-1)*t, res+=d*t;
}
if(m) printf("-1\n");
else printf("%lld\n", res);
}
}
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78B - Easter Eggs | 1455B - Jumps |
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1257A - Two Rival Students | 1415A - Prison Break |
1271A - Suits | 259B - Little Elephant and Magic Square |
1389A - LCM Problem | 778A - String Game |
1382A - Common Subsequence | 1512D - Corrupted Array |
667B - Coat of Anticubism | 284B - Cows and Poker Game |
1666D - Deletive Editing | 1433D - Districts Connection |
2B - The least round way | 1324A - Yet Another Tetris Problem |
246B - Increase and Decrease | 22E - Scheme |
1566A - Median Maximization | 1278A - Shuffle Hashing |