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题意：给你n 个数字，某一个数的幸运数是这个数前面比他小 离他最远的位置之差，求出最大幸运值。
析：先按从大到小排序，然后去维护那个最大的id，一直比较，更新最大值就好。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 5;
const int mod = 1e8;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, -1, 1, 1, -1};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
struct node{
int x, id;
bool operator < (const node &p) const{
return x > p.x;
}
};
node a[maxn];
int main(){
int T; cin >> T;
while(T--){
scanf("%d", &n);
for(int i = 0; i < n; ++i){
scanf("%d", &a[i].x);
a[i].id = i;
}
sort(a, a+n);
int ans = 0, mmax = -1, tmp = -1;
for(int i = 0; i < n; ++i){
ans = max(ans, mmax-a[i].id);
tmp = max(tmp, a[i].id);
if(i != n-1 && a[i].x > a[i+1].x)
mmax = max(tmp, mmax);
}
printf("%d\n", ans);
}
return 0;
}
UVALive 6692 Lucky Number （思路 + 枚举）
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原文地址：http://www.cnblogs.com/dwtfukgv/p/5800867.html
