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  • Grand Garden思维题

    题目描述

    In a flower bed, there are N flowers, numbered 1,2,…,N. Initially, the heights of all flowers are 0. You are given a sequence h={h1,h2,h3,…} as input. You would like to change the height of Flower k to hk for all k (1≤k≤N), by repeating the following “watering” operation:
    Specify integers l and r. Increase the height of Flower x by
    1 for all x such that l≤x≤r.
    Find the minimum number of watering operations required to satisfy the condition.

    Constraints
    ·1≤N≤1000≤hi≤100
    ·All values in input are integers.

    输入

    Input is given from Standard Input in the following format:
    N
    h1 h2 h3 … hN

    输出

    Print the minimum number of watering operations required to satisfy the condition.

    样例输入

    4
    1 2 2 1
    

    样例输出

    2

    提示

    The minimum number of watering operations required is 2. One way to achieve it is:
    Perform the operation with (l,r)=(1,3).
    Perform the operation with (l,r)=(2,4).

    这是一道思维题

    #pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
    #pragma GCC optimize("Ofast")
    #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
    #pragma comment(linker, "/stack:200000000")
    #pragma GCC optimize (2)
    #pragma G++ optimize (2)
    #include <bits/stdc++.h>
    #include <algorithm>
    #include <map>
    #include <queue>
    #include <set>
    #include <stack>
    #include <string>
    #include <vector>
    using namespace std;
    #define wuyt main
    typedef long long ll;
    #define HEAP(...) priority_queue<__VA_ARGS__ >
    #define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > >
    template<class T> inline T min(T &x,const T &y){return x>y?y:x;}
    template<class T> inline T max(T &x,const T &y){return x<y?y:x;}
    //#define getchar()(p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
    //char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf;
    ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar();
    if(c == '-')Nig = -1,c = getchar();
    while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar();
    return Nig*x;}
    #define read read()
    const ll inf = 1e15;
    const int maxn = 3e5 + 10;
    const int mod = 1e9 + 7;
    #define start int wuyt()
    #define end return 0
    ll num[maxn],ans;
    ll n,m;
    ll a[maxn],s[maxn];
    start{
        int n=read;
        for(int i=1;i<=n;i++) num[i]=read;
        for(int i=1;i<=n;i++)
            if(num[i-1]<num[i]) 
            	ans+=num[i]-num[i-1];
        cout<<ans;
    	end;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/PushyTao/p/13144217.html
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