[Zhengrui 2021 winter vacation provincial election second round of training day 1] Token generation (number of combinations + two points)

description

solution

The number of yyds that meet the conditions is actually the same as $n$ (not the meaning of the title) is inextricably linked to the arithmetic sequence formula,
so the specific value can be divided into two
. The value range of the final answer must grow into$[,)$ , the left is closed and the right is opened,
and the two boundaries must be the one with the smallest difference$1$ , then the binary aspect of the answer is also fixed and the

same is divided into1 1The specific position of$1$ can be
linked here with the number of combinations, which can preprocess a small number of combinations to reduce query time complexity

Privately think that it is still difficult to think!

code

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
#define MAX 1e18
#define ll long long
#define mod 998244353
ll c[3005][3005];

ll qkpow1( ll x, ll y ) {
    
    
	ll ans = 1;
	while( y ) {
    
    
		if( y & 1 ) ans = ans * x % mod;
		x = x * x % mod;
		y >>= 1;
	}
	return ans;
}

ll qkpow2( ll x, ll y ) {
    
    
	ll ans = 1;
	while( y ) {
    
    
		if( y & 1 ) ans *= x;
		x *= x;
		y >>= 1;
	}
	return ans;
}

ll calc( ll n ) {
    
    
	return n * ( n + 1 ) >> 1;
}

ll C( ll n, ll m ) {
    
    
	if( n <= 3000 && m <= 3000 ) return c[n][m];
	if( m > n - m ) m = n - m;
	if( n >= 1e7 && m >= 3 ) return MAX + 1;
	__int128 d1 = 1, d2 = 1;
	for( int i = 1;i <= m;i ++ ) d1 *= i;
	for( int i = n - m + 1;i <= n;i ++ ) {
    
    
		d2 *= i;
		if( d2 > d1 * MAX ) return MAX + 1;
	}
	return d2 / d1;
}

void solve( ll n, ll k, ll ans ) {
    
    
	ll t = 0, tmp;
	while( k > ( tmp = C( n, t ) ) ) k -= tmp, t ++;
	int last = n;
	for( ll i = 1;i <= t;i ++ ) {
    
    
		ll l = 1, r = last, pos;
		while( l <= r ) {
    
    
			ll mid = ( l + r ) >> 1;
			if( C( mid - 1, t - i + 1 ) < k ) pos = mid, l = mid + 1;
			else r = mid - 1;
		}
		last = pos - 1;
		k -= C( pos - 1, t - i + 1 );
		ans = ( ans + qkpow1( 2, pos - 1 ) ) % mod;
	}
	printf( "%lld\n", ans );
}

void init( int n = 3000 ) {
    
    
	for( int i = 0;i <= n;i ++ )
		for( int j = 1;j <= i;j ++ )
			c[i][j] = MAX + 1;
	for( int i = 0;i <= n;i ++ ) {
    
    
		c[i][0] = 1;
		for( int j = 1;j <= i;j ++ )
			c[i][j] = min( c[i][j], c[i - 1][j] + c[i - 1][j - 1] );
	}
}

int main() {
    
    
	init();
	int T; ll n, k;
	scanf( "%d", &T );
	while( T -- ) {
    
    
		scanf( "%lld %lld", &n, &k );
		ll t = sqrt( n << 1 );
		while( calc( t ) > n ) t --;
		ll t1 = n - calc( t ), t2 = t - t1;
		if( ! t1 ) {
    
    
			if( k == 1 ) printf( "%lld\n", ( qkpow1( 2, t ) - 1 + mod ) % mod );
			else printf( "-1\n" );
		}
		else if( t2 < 60 && qkpow2( 2, t2 ) < k )
			printf( "-1\n" );
		else solve( t2, k, ( qkpow1( 2, t1 ) - 1 + mod ) % mod * qkpow1( 2, t2 + 1 ) % mod );
	}
	return 0;
}