标签：des os io ar for div cti sp amp
Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (n?+?1) pylons numbered from 0 to n in this game. The pylon with number 0 has zero height, the pylon with numberi (i?>?0) has heighthi. The goal of the game is to reachn-th pylon, and the only move the player can do is to jump from the current pylon (let‘s denote its number ask) to the next one (its number will be k?+?1). When the player have made such a move, its energy increases byhk?-?hk?+?1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn‘t want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
The first line contains integer n (1?≤?n?≤?105). The next line containsn integers h1,h2,?..., hn (1??≤??hi??≤??105) representing the heights of the pylons.
Print a single number representing the minimum number of dollars paid by Caisa.
5 3 4 3 2 4
4
3 4 4 4
4
In the first sample he can pay 4 dollars and increase the height of pylon with number0 by 4 units. Then he can safely pass to the last pylon.
题意：依次走过n+1个地方，每次从第i个走到第i+1个都会有hi-hi+1的能量，正的话，是加血，负的话是扣血，为了确保玩这个游戏，所以要保证血是正的，也可以通过花钱来增加高度来抵消血的损失，求最少的花费
思路：单纯的模拟过去
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
typedef long long ll;
using namespace std;
const int maxn = 100005;
int arr[maxn];
int brr[maxn];
int main() {
ll ans = 0;
int n, a;
scanf("%d", &n);
int cnt = 0;
for (int i = 1; i <= n; i++) {
scanf("%d", &a);
if (cnt < a) {
ans += a - cnt;
cnt = a;
}
}
cout << ans << endl;
return 0;
}codeforces B#264. Caisa and Pylons
标签：des os io ar for div cti sp amp
原文地址：http://blog.csdn.net/u011345136/article/details/38947397
