Vanya and Scales CodeForces - 552C (思维)

大意: $101$个砝码, 重$w^0,w^1,...,w^100$, 求能否称出重量$m$.

w<=3时显然可以称出所有重量, 否则可以暴力双端搜索.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <unordered_map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head




int w, m;
ll a[100];
unordered_map<ll,int> f[2];
void dfs(int d, int mx, ll num, unordered_map<ll,int>& f) {
	if (d>mx) ++f[num];
	else { 
		dfs(d+1,mx,num,f);
		dfs(d+1,mx,num+a[d],f);
		dfs(d+1,mx,num-a[d],f);
	}
}

int main() {
	scanf("%d%d", &w, &m);
	if (w<=3) return puts("YES"),0;
	ll now = 1;
	REP(i,0,16) {
		a[++*a] = now;
		now *= w;
	}
	dfs(1,*a/2,0,f[0]);
	dfs(*a/2+1,*a,0,f[1]);
	for (auto &&t:f[0]) {
		if (f[1].count(t.x+m)) return puts("YES"),0;
	}
	puts("NO");
}

实际上有更优做法.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <unordered_map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head





int main() {
	int w, m;
	scanf("%d%d", &w, &m);
	while (m) {
		if ((m-1)%w==0) --m;
		else if ((m+1)%w==0) ++m;
		else if (m%w) return puts("NO"),0;
		m /= w;
	}
	puts("YES");
}