zoukankan      html  css  js  c++  java
  • CF div2 331 B

    B. Wilbur and Array
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Wilbur the pig is tinkering with arrays again. He has the array a1, a2, ..., an initially consisting of n zeros. At one step, he can choose any index i and either add 1 to all elements ai, ai + 1, ... , an or subtract 1 from all elements ai, ai + 1, ..., an. His goal is to end up with the array b1, b2, ..., bn.

    Of course, Wilbur wants to achieve this goal in the minimum number of steps and asks you to compute this value.

    Input

    The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the length of the array ai. Initially ai = 0 for every position i, so this array is not given in the input.

    The second line of the input contains n integers b1, b2, ..., bn ( - 109 ≤ bi ≤ 109).

    Output

    Print the minimum number of steps that Wilbur needs to make in order to achieve ai = bi for all i.

    Sample test(s)
    input
    5
    1 2 3 4 5
    output
    5
    input
    4
    1 2 2 1
    output
    3
    Note

    In the first sample, Wilbur may successively choose indices 1, 2, 3, 4, and 5, and add 1 to corresponding suffixes.

    In the second sample, Wilbur first chooses indices 1 and 2 and adds 1 to corresponding suffixes, then he chooses index 4 and subtract1.

    也是比较水,模拟就跑一遍就行了,然后用long long 

    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #include <stack>
    #include <queue>
    #include <map>
    #include <algorithm>
    #include <vector>
    
    using namespace std;
    
    const int maxn = 200005;
    
    typedef long long LL;
    
    vector<int>G[maxn];
    
    int a[maxn];
    int b[maxn];
    
    
    
    int main()
    {
        int n,m;
        int tmp = 0;
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d",&b[i]);
        }
        LL ans = 0;
        int add = 0,sub = 0;
        memset(a,0,sizeof(a));
        for(int i = 1;i <= n;i++){
            if(add) a[i] += add;
            if(sub) a[i] -= sub;
            if(a[i]!=b[i]){
                if(a[i]>b[i]) {sub += (a[i]-b[i]);ans += (a[i]-b[i]);}
                else { add += (b[i]-a[i]);ans += (b[i]-a[i]);}
            }
        }
    
    
        printf("%lld
    ",ans);
    
    
        return 0;
    }
    View Code
  • 相关阅读:
    OBJ文件格式详解
    HashMap的用法
    HashMap和Hashtable的区别
    加载物体的方法
    drawSelf(int texId)格式对应
    adb.exe诊断
    Android Eclipse如何用BlueStacks模拟器
    .md5mesh and .md5anim文件介绍
    ubuntu命令行下中文乱码的解决方案 (我采取了其中方案一与方案二,都还没成功—待定)
    Ubuntu下小巧智能的代码编辑器Scribes
  • 原文地址:https://www.cnblogs.com/lmlyzxiao/p/4979572.html
Copyright © 2011-2022 走看看