Back Gear | Block Contest

Title Description Description

Spring and Spring Kindergarten held the annual "Building Block Contest". This year's game is to build a building with a width of n, the building can be seen by the n block width of 1 blocks, the final height of block I need to be hi.

Before the construction begins, there is no building block (it can be seen as a block of n blocks of height 0). After each operation, the children can select a continuous interval [L, r] and then increase the height of all blocks from block L to block R (including Block L and R Block) by 1 respectively.

Little M was a clever little friend, and she soon came up with the best strategy for building, making the minimum number of operations required. But she is not a diligent child, so I would like to ask you to help achieve this strategy, and to find the minimum number of operations.

Input/output format Input/output

Input format:

Input file is block.in

The input contains two lines, and the first row contains an integer n, which indicates the width of the building.

The second line contains n integers, and the first integer is hi.

Output format:

Output file is Block.out

Only one row, which is the minimum number of operations required to build.

Input and Output Sample sample Input/output

Sample test point # # Input Example:

5

2 3 4) 1 2

Sample output:

5

Description description

"Sample Interpretation"

One of the best possible scenarios, select

[1,5] [1,3] [2,3] [3,3] [5,5]

"Data Range"

For 30% of the data, there are 1≤n≤10;

For 70% of the data, there are 1≤n≤1000;

For 100% of the data, there is 1≤n≤100000,0≤hi≤10000.

Topic Analysis:

The subject surface is a simulation, in fact, is a simulation ... But this problem online many Daniel have adopted the NLOGN algorithm ... is the online data too weak? The O (n) code solves this problem and has been evaluated by TYVJ and Rokua. (In the beginning, I forgot to add Cstdio library, wasted two times, my pass rate ah ... ）

#include <iostream>#include<cstdio>using namespacestd;intMain () {intn,now=0, ans=0; scanf ("%d",&N);  for(intI=1; i<=n; i++) {  intT; scanf ("%d",&t); if(T>now) ans+=t-Now ; now=T;} cout<<ans;return 0;}

Back Gear | Block Contest