zoukankan      html  css  js  c++  java
  • E. Mahmoud and Ehab and the function Codeforces Round #435 (Div. 2)

    http://codeforces.com/contest/862/problem/E

    二分答案

    一个数与数组中的哪个数最接近:

    先对数组中的数排序,然后lower_bound

     1 #include <cstdio>
     2 #include <cstdlib>
     3 #include <cmath>
     4 #include <cstring>
     5 #include <time.h>
     6 #include <string>
     7 #include <set>
     8 #include <map>
     9 #include <list>
    10 #include <stack>
    11 #include <queue>
    12 #include <vector>
    13 #include <bitset>
    14 #include <ext/rope>
    15 #include <algorithm>
    16 #include <iostream>
    17 using namespace std;
    18 #define ll long long
    19 #define minv 1e-6
    20 #define inf 1e18
    21 #define pi 3.1415926536
    22 #define E  2.7182818284
    23 const ll mod=1e9+7;//998244353
    24 const int maxn=1e5+10;
    25 
    26 ll b[maxn],f[maxn];
    27 
    28 int main()
    29 {
    30     int n,m,q,g=0,i,l,r;
    31     ll a,tot=0,sum=0,v;
    32     scanf("%d%d%d",&n,&m,&q);
    33     for (i=1;i<=n;i++)
    34     {
    35         scanf("%lld",&a);
    36         if (i & 1)
    37             sum+=a;
    38         else
    39             sum-=a;
    40     }
    41     for (i=1;i<=m;i++)
    42     {
    43         scanf("%lld",&b[i]);
    44         if (i & 1)
    45             tot+=b[i];
    46         else
    47             tot-=b[i];
    48         if (i>=n)
    49         {
    50             g++;
    51             if ((i-n) & 1)
    52                 f[g]=-tot;
    53             else
    54                 f[g]=tot;
    55             if ((i-n+1) & 1)
    56                 tot-=b[i-n+1];
    57             else
    58                 tot+=b[i-n+1];
    59         }
    60     }
    61     sort(f+1,f+g+1);
    62 
    63         i=lower_bound(f+1,f+g+1,sum)-f;
    64         v=inf;
    65         if (i-1>=1 && i-1<=g)
    66             v=min(v,abs(sum-f[i-1]));
    67         if (i>=1 && i<=g)
    68             v=min(v,abs(sum-f[i]));
    69         if (i+1>=1 && i+1<=g)
    70             v=min(v,abs(sum-f[i+1]));
    71         printf("%lld
    ",v);
    72 
    73     while (q--)
    74     {
    75         scanf("%d%d%lld",&l,&r,&a);
    76         if ((r-l)%2==0)
    77         {
    78             if (l & 1)
    79                 sum+=a;
    80             else
    81                 sum-=a;
    82         }
    83 
    84         i=lower_bound(f+1,f+g+1,sum)-f;
    85         v=inf;
    86         if (i-1>=1 && i-1<=g)
    87             v=min(v,abs(sum-f[i-1]));
    88         if (i>=1 && i<=g)
    89             v=min(v,abs(sum-f[i]));
    90         if (i+1>=1 && i+1<=g)
    91             v=min(v,abs(sum-f[i+1]));
    92         printf("%lld
    ",v);
    93     }
    94     return 0;
    95 }
  • 相关阅读:
    设计模式相关,单例模式!
    混合App交互相关,jsBridge的原理
    Vue相关,插槽怎么用!
    Mysql相关,查询功能整理
    图文并茂讲解进程与线程
    线程与进程的区别及其通信方式
    算法,十进制数转任意进制数
    JavaScript,leetcode第198题,打家劫舍
    2.4G无线控制器附加AT2401C功放IC增加距离
    超低功耗WiFi :ESP8089
  • 原文地址:https://www.cnblogs.com/cmyg/p/9520945.html
Copyright © 2011-2022 走看看