zoukankan      html  css  js  c++  java
  • 2015 多校联赛 ——HDU5353(构造)

    Each soda has some candies in their hand. And they want to make the number of candies the same by doing some taking and giving operations. More specifically, every two adjacent soda x and y can do one of the following operations only once:
    1. x-th soda gives y-th soda a candy if he has one;
    2. y-th soda gives x-th soda a candy if he has one;
    3. they just do nothing.

    Sample Input
    3 6 1 0 1 0 0 0 5 1 1 1 1 1 3 1 2 3
     
    Sample Output
    NO YES 0 YES 2 2 1 3 2


    对相邻两个数之间进行以下三种操作的一种,最后使他们相等

    ①a++   b--         ②a--   b++      ③nothing  

    如果(sum%n != 0),直接失败。用一个数组来记录数与平均数之间的差值。

    先枚举第一位数的三种情况,

    当C[i] == 1时,从C[i+1]取一;

    当C[i] == -1时,给C[i+1]一个;

    当C[i] == 0时,nothing;

    else:false。

    ps:完全没想到要对第一位进行枚举 OoO,一直wa


    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <vector>
    #pragma comment(linker, "/STACK:102400000,102400000")
    using namespace std;
    typedef long long ll;
    const int mod = 1000000007;
    int a[100050];
    int aver,flag,n;
    int p[100050][2];
    int c[100050];
    int tot;
    
    bool solve()
    {
        for(int i = 3; i <= n; i++)
            c[i] = a[i];
        for(int i = 2; i <= n; i++)
        {
            if(c[i] == 0)
                continue;
            else if(c[i] == 1 && i != n)
            {
                c[i+1]++;
                c[i]--;
                p[tot][0] = i;
                p[tot++][1] = i+1;
            }
            else if(c[i] == 1 && i == n)
            {
                c[1]++;
                c[i]--;
                p[tot][0] = i;
                p[tot++][1] = 1;
            }
            else if(c[i] == -1 && i!= n)
            {
                c[i]++;
                c[i+1]--;
                p[tot][0] = i+1;
                p[tot++][1] = i;
            }
            else if(c[i] == -1 && i== n)
            {
                c[i]++;
                c[1]--;
                p[tot][0] = 1;
                p[tot++][1] = i;
            }
            else if(c[i] >1 || c[i] < -1)
                return false;
        }
        for(int i = 1; i <= n; i++)
            if(c[i]!=0)
                return false;
        return true;
    }
    
    void prin()
    {
        printf("YES
    ");
        printf("%d
    ",tot);
        for(int i = 0; i < tot; i++)
            printf("%d %d
    ",p[i][0],p[i][1]);
    }
    
    int main()
    {
        int T;
        //freopen("01.txt","r",stdin);
        scanf("%d",&T);
        while(T--)
        {
            scanf("%d",&n);
            ll sum = 0;
            memset(p,0,sizeof(p));
            for(int i = 1; i <= n; i++)
            {
                scanf("%d",&a[i]);
                sum += a[i];
            }
    
            if(sum % n)
            {
                printf("NO
    ");
                continue;
            }
            aver = sum / n;
            flag = 0;
            tot = 0;
            for(int j = 1; j <= n; j++)
                a[j] =a[j] - aver;
            c[1] = a[1];
            c[2] = a[2];
            if(solve())
            {
                prin();
            }
            else
            {
                tot = 0;
                c[1] = a[1] - 1;
                c[2] = a[2] + 1;
                p[tot][0] = 1;
                p[tot++][1] = 2;
                if(solve())
                    prin();
                else
                {
                    tot = 0;
                    c[1] = a[1] + 1;
                    c[2] = a[2] - 1;
                    p[tot][0] = 2;
                    p[tot++][1] = 1;
                    if(solve())
                        prin();
                    else
                        printf("NO
    ");
                }
            }
    
        }
        return 0;
    }
    

      

  • 相关阅读:
    Vue-router的实现原理
    get请求被浏览器跨域的同源策略请求机制拦截,但是get请求是否请求到了服务器呢
    合并两个有序链表
    JS实现链式调用 a().b().c()
    CSS知识点总结
    BK-信息查找、摘取
    radar图生成用户guideline
    【转】 mybatis 详解(七)------一对一、一对多、多对多
    【转】 mybatis 详解(六)------通过mapper接口加载映射文件
    【转】 mybatis 详解(五)------动态SQL
  • 原文地址:https://www.cnblogs.com/Przz/p/5409798.html
Copyright © 2011-2022 走看看