Note
Note to the first sample test. One pan can have an item of mass7and a weight of mass3, and the second pan can have two weights of masses9and1, correspondingly. Then7+3=9+1.
Note to the second sample test. One pan of the scales can have an item of mass99and the weight of mass1, and the second pan can have the weight of mass100.
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 1000000007int 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;}
生命不息，奋斗不止，就像我们常说的一句话；
