UVA - 11078 Open Credit System 推理

题目大意：给一个长度为n的整数序列A0，A1，A2.。。An-1，找出两个整数Ai和Aj（i<j），使得Ai-Aj尽量大

解题思路：用MAX记录最大的Ai，MIN记录最小的Aj

当num[i]大于MAX时，令MAX = Aj，MIN=INF

当num[i] <MIN时，MIN = num[i],令ans = min(ans,MAX-MIN)

#include<algorithm>#include<cstdio>#include<cstring>using namespace std;#define INF 0x3f3f3f3f#define maxn 100010int num[maxn];int main() {int test, n, x;scanf("%d", &test);while(test--) {scanf("%d",&n);for(int i = 0; i < n; i++) scanf("%d", &num[i]);int ans = num[0] - num[1], MAX = num[0], MIN = num[1];for(int i = 2; i < n; i++) {if(num[i] > MAX) {MAX = num[i];MIN = INF;continue;}if(num[i] < MIN) { MIN = num[i];ans = max(MAX-MIN,ans);}}printf("%d\n",ans);}return 0;}