例题1.18 开放式学分制 UVa11078

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1.题目描述：点击打开链接

2.解题思路：本题一看n的范围高达100000，肯定只能用O(N)的复杂度解决。本题类似于最大连续和问题，事先计算区间[0,i)的最大值，存放到_max数组中，然后扫描整个数组，不断用max(ans,_max[i]-a[i])更新最大差值即可。本题附上利用O(1)更新MaxAi来计算最大值的程序。更新顺序值得学习！

3.代码：

#define _CRT_SECURE_NO_WARNINGS #include<iostream>#include<algorithm>#include<string>#include<sstream>#include<set>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<functional>using namespace std;#define N 100000+10#define INF 150000+10int a[N];int _max[N];void init(){for (int i = 0; i < N; i++)_max[i] = -INF;}int main(){//freopen("t.txt", "r", stdin);int T;cin >> T;while (T--){init();int n;cin >> n;for (int i = 0; i < n; i++)cin >> a[i];for (int i = 1; i <= n; i++)_max[i] = max(_max[i - 1], a[i - 1]);int ans = -2 * INF;for (int i = 1; i < n; i++)ans = max(ans, _max[i] - a[i]);printf("%d\n", ans);}return 0;}

参考程序：

#define _CRT_SECURE_NO_WARNINGS #include<iostream>#include<algorithm>#include<string>#include<sstream>#include<set>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<functional>using namespace std;#define N 100000int a[N], n;int main(){//freopen("t.txt", "r", stdin);int T;cin >> T;while (T--){scanf("%d", &n);for (int i = 0; i < n; i++)scanf("%d", a + i);int ans = a[0] - a[1];int MaxAi = a[0];for (int j = 1; j < n; j++){ans = max(ans, MaxAi - a[j]);MaxAi = max(a[j], MaxAi);//MaxAi晚于ans更新，因为更新ans时候MaxAi不包括第j个数}printf("%d\n", ans);}return 0;}

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