Codeforces Round #308 (Div. 2) C. Vanya and Scales

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Vanya has a scales for weighing loads and weights of masses w0, w1, w2, ..., w100 grams where w is some integer not less than 2(exactly one weight of each nominal value). Vanya wonders whether he can weight an item with mass m using the given weights, if the weights can be put on both pans of the scales. Formally speaking, your task is to determine whether it is possible to place an item of mass m and some weights on the left pan of the scales, and some weights on the right pan of the scales so that the pans of the scales were in balance.

Input

The first line contains two integers w, m (2 ≤ w ≤ 109, 1 ≤ m ≤ 109) — the number defining the masses of the weights and the mass of the item.

Output

Print word 'YES' if the item can be weighted and 'NO' if it cannot.

Sample test(s) 
input 
3 7
output 
YES
input 
100 99
output 
YES
input 
100 50
output 
NO

Note

Note to the first sample test. One pan can have an item of mass 7 and a weight of mass 3, and the second pan can have two weights of masses 9 and 1, correspondingly. Then 7 + 3 = 9 + 1.

Note to the second sample test. One pan of the scales can have an item of mass 99 and the weight of mass 1, and the second pan can have the weight of mass 100.

Note to the third sample test. It is impossible to measure the weight of the item in the manner described in the input.

题目要求把一个m分解成w进制，每个进制位上要求是0 1 -1这三个数，用公式 (w-1)w^(n-1) = w^n - w^(n-1);可得当一个位上是w-1时，可以向上一位

借一个数，使本位为-1,这样就符合要求了，很最低位向最高位一个个改就可以了，改不了，就输出no,复杂度为logn.其实，马上可以得到当w = 2,3时，任何数字都是可以构造出来的，这样，也可以直接用暴搜，当w很大时，分解的位数不是很多，所以很快能搜到，当w = 2,3直接得到答案就可以了，这样复杂度明显高了很多倍了！

#define N 1005
#define MOD 1000000007
int n,w,m,mIndex,pri[100];
long long num[100];
bool Dfs(long long goal,long long now,int index){if(now > goal)return false;if(now == goal)return true;if(index > mIndex)return false;if(Dfs(goal ,now + num[index],index+1))return true;if(Dfs(goal ,now - num[index],index+1))return true;if(Dfs(goal ,now ,index+1))return true;return false;
}
void solve1(){if(w <= 3){printf("YES\n");return ;}num[0] = 1;for(int i=1;i<100;i++){num[i] = num[i-1] * w;mIndex = i;if(num[i] >= MOD)break;}if(Dfs(m,0,0)){printf("YES\n");}elseprintf("NO\n");
}
void solve2(){int index = 0;while(m){pri[index++] = m % w;m = m/w;}FI(index){//printf("index %d %d \n",i,pri[i]);if(pri[i]!=0  && pri[i]!= 1 && pri[i] != w && pri[i] != (w -1)){printf("NO\n");return ;}if(pri[i] == w-1 || pri[i] == w){pri[i+1]++;}}printf("YES\n");
}
int main()
{while (S2(w,m) != EOF){solve1();}return 0;
}

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