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; }
