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23 lines
669 B
23 lines
669 B
#include <bits/stdc++.h>
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using namespace std;
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const int N = 1000010;
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//递推法
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int n, a[N];
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int f[N]; //代表前i个区域被填充好的最少次数.套路，满满的套路
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int main() {
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cin >> n;
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for (int i = 1; i <= n; i++) cin >> a[i];
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//递推出口，第一个有多大的坑，不能指望别人，都要靠自己来完成~
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f[1] = a[1];
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//从第2个开始进行递推
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for (int i = 2; i <= n; i++) {
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//递推式：见题解的分析过程
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if (a[i] <= a[i - 1]) f[i] = f[i - 1];
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else f[i] = f[i - 1] + (a[i] - a[i - 1]);
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}
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//输出大吉
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cout << f[n] << endl;
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return 0;
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} |