UVa 11549 Open Credit System

题意：给出n个数，找出两个整数a[i],a[j](i < j),使得a[i] - a[j]尽量大

从小到大枚举j,在这个过程中维护a[i]的最大值

maxai晚于ans更新，

可以看这个例子

 1 8 9 10 11

正确的应该是-1

如果更早更新的话，算出来就是0

用数组来存的

#include<iostream>  
#include<cstdio>  
#include<cstring> 
#include <cmath> 
#include<stack>
#include<vector>
#include<map> 
#include<set>
#include<queue> 
#include<algorithm>  
using namespace std;

typedef long long LL;
const int INF = (1<<30)-1;
const int mod=1000000007;
const int maxn=1000005;

int a[maxn];
int 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("j = %d  maxai = %d  ans = %d
",j,maxai,ans);
        }
        printf("%d
",ans);
    }
    return 0;
}

边读边算的

#include<iostream>  
#include<cstdio>  
#include<cstring> 
#include <cmath> 
#include<stack>
#include<vector>
#include<map> 
#include<set>
#include<queue> 
#include<algorithm>  
using namespace std;

typedef long long LL;
const int INF = (1<<30)-1;
const int mod=1000000007;
const int maxn=1000005;

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        int n;
        scanf("%d",&n);

        int x,y,ans;
        scanf("%d %d",&x,&y);
        ans = x-y;
        int maxai = x;
        for(int i = 2;i<n;i++){
            scanf("%d",&x);
            ans = max(ans,maxai - x);
            maxai = max(maxai,x);
        }
        printf("%d
",ans);
    }
    return 0;
}