zoukankan      html  css  js  c++  java
  • CodeForces D. Walking Between Houses

    http://codeforces.com/contest/1015/problem/D

    There are nn houses in a row. They are numbered from 11 to nn in order from left to right. Initially you are in the house 11.

    You have to perform kk moves to other house. In one move you go from your current house to some other house. You can't stay where you are (i.e., in each move the new house differs from the current house). If you go from the house xx to the house yy, the total distance you walked increases by |xy||x−y| units of distance, where |a||a| is the absolute value of aa. It is possible to visit the same house multiple times (but you can't visit the same house in sequence).

    Your goal is to walk exactly ss units of distance in total.

    If it is impossible, print "NO". Otherwise print "YES" and any of the ways to do that. Remember that you should do exactly kk moves.

    Input

    The first line of the input contains three integers nn, kk, ss (2n1092≤n≤109, 1k21051≤k≤2⋅105, 1s10181≤s≤1018) — the number of houses, the number of moves and the total distance you want to walk.

    Output

    If you cannot perform kk moves with total walking distance equal to ss, print "NO".

    Otherwise print "YES" on the first line and then print exactly kk integers hihi (1hin1≤hi≤n) on the second line, where hihi is the house you visit on the ii-th move.

    For each jj from 11 to k1k−1 the following condition should be satisfied: hjhj+1hj≠hj+1. Also h11h1≠1 should be satisfied.

    Examples
    input
    Copy
    10 2 15
    output
    Copy
    YES
    10 4
    input
    Copy
    10 9 45
    output
    Copy
    YES
    10 1 10 1 2 1 2 1 6
    input
    Copy
    10 9 81
    output
    Copy
    YES
    10 1 10 1 10 1 10 1 10
    input
    Copy
    10 9 82
    output
    Copy
    NO

    代码:

    #include <bits/stdc++.h>
    using namespace std;
    
    typedef long long ll;
    ll N, K, S;
    
    int main() {
        cin >>N >> K >> S;
        if((N - 1) * K < S || S < K) {
            printf("NO
    ");
            return 0;
        }
    
        printf("YES
    ");
        int st = 1;
        while(K --) {
            int dis = min(S - K, N - 1);
            S -= dis;
            if(st + dis <= N) st += dis;
            else st -= dis;
            printf("%d ", st);
        }
        printf("
    ");
        return 0;
    }
    

      

  • 相关阅读:
    SQL server 数据库基础语句
    数据库学习的第一天
    C# 函数
    C# for循环的嵌套 作用域
    C# for循环语句
    Docker的基本使用
    django连接postgresql
    docker的安装
    Postgresql的使用
    Celery的介绍
  • 原文地址:https://www.cnblogs.com/zlrrrr/p/9858444.html
Copyright © 2011-2022 走看看