Codeforces 552C. Vanya and Scales【巧】
2016-04-22 21:45
369 查看
C. Vanya and Scales
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
Examples
input
output
input
output
input
output
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.
恩,砝码,放天平上,问是否能使天平平衡,天平两侧都可放砝码,每个砝码只能用一次。比较巧妙的想法,将m 换成w进制,因为砝码w^0 ,w^1 ,w^2 ,w^3等等即1 ,10 ,100,1000等等,所以只有换成w 进制之后某位数值为w-1才可加1 ,而为0 或1 的不用处理,为其他数值则要输出NO
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
Examples
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.
恩,砝码,放天平上,问是否能使天平平衡,天平两侧都可放砝码,每个砝码只能用一次。比较巧妙的想法,将m 换成w进制,因为砝码w^0 ,w^1 ,w^2 ,w^3等等即1 ,10 ,100,1000等等,所以只有换成w 进制之后某位数值为w-1才可加1 ,而为0 或1 的不用处理,为其他数值则要输出NO
#include <iostream> #include<cstdio> #include<cstring> #include<cmath> using namespace std; int r[1010]; int main() { int w,m; while(~scanf("%d%d",&w,&m)) { int k=0; while(m) { r[k++]=m%w; m=m/w; } int i; for(i=0;i<k;++i) { if(r[i]>=w)//开始没想到处理这种情况,看错误数据,调试一遍才找到 { r[i+1]++; r[i]=r[i]%w; } if(r[i]<=1) continue; else if(r[i]==w-1) r[i]=0,r[i+1]++; else break; } if(i==k) printf("YES\n"); else printf("NO\n"); } return 0; }
相关文章推荐
- 站立会议第五天
- 团队项目个人每日总结(4.22)
- 典型用户分析
- Android - View的绘制流程二(layout)
- 【Raspberry Pi 3试用体验】+ 中文显示及输入+百度云传输
- Shell中创建序列和数组(list、array)的方法
- HDU 5671 矩阵的交换行、交换列,整行加一个数 思维题
- iOS基础:NSDate
- Android stadio
- iPhone4s刷机教程
- Android Studio2.0在Android5.0以下机型无法调试
- vb.net操作excel时,如何判断单元格内容为空
- Windows下使用Beyond Compare作为git的比对与合并工具
- JDBC01
- 冲刺第二天
- 转jmeter --JDBC请求
- 算法导论:堆排序
- hdu 5667 矩阵快速幂
- hdoj1008
- BNUOJ-4716题 IQ test