UVA - 11078 Open Credit System
2013-11-30 18:31
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题意:求出数列中差值最大的数是多少,O(n)的算法是:因为它要求i<j,求(Ai-Aj)所以其实当我们遍历每一个数的时候,可以用一个值记录之前的最大值
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int A[100000],n; int main(){ int t; scanf("%d",&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(MaxAi,A[j]); } printf("%d\n",ans); } return 0; }
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