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  • Vanya and Scales(思维)

    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
    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.

    题解:很棒的一个题,初一看完全没思路,到最后也没搞定。。。这个题的意思是给w的(0-100)次方,0-100次方只用一次,使天平平衡,

    转化一下就是一个w进制的数能否通过加减w的(0-100)次方,0-100次方只用一次,使之等于0;

    代码:

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<cmath>
    #include<algorithm>
    using namespace std;
    int a[110];
    int main(){
        int w, m;
        while(~scanf("%d%d", &w, &m)){
            int k = 0; 
            while(m){
                a[k++] = m % w;
                m /= w;
            }
            int ans = 1;
            for(int i = 0; i < k; i++){
                if(a[i] == w){
                    a[i] = 0;
                    a[i + 1]++;
                    continue;
                }
                if(a[i] <= 1){
                    continue;
                }
                else if(a[i] == w - 1){
                    a[i + 1]++;
                    a[i] = 0;
                    continue;
                }
                else
                    ans = 0;
            }
            if(ans)
                puts("YES");
            else
                puts("NO");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/handsomecui/p/5418468.html
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