题目链接：http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1272
思路：排序后由于序列都是顺序的，则只需要考虑序号了，加入当前维护的序号比后面的小，则更新ret。否则更新当前序号
1 #include <algorithm> 2 #include <iostream> 3 #include <iomanip> 4 #include <cstring> 5 #include <climits> 6 #include <complex> 7 #include <fstream> 8 #include <cassert> 9 #include <cstdio> 10 #include <bitset> 11 #include <vector> 12 #include <deque> 13 #include <queue> 14 #include <stack> 15 #include <ctime> 16 #include <set> 17 #include <map> 18 #include <cmath> 19 20 using namespace std; 21 #define fr first 22 #define sc second 23 #define cl clear 24 #define BUG puts("here!!!") 25 #define W(a) while(a--) 26 #define pb(a) push_back(a) 27 #define Rint(a) scanf("%d", &a) 28 #define Rll(a) scanf("%lld", &a) 29 #define Rs(a) scanf("%s", a) 30 #define Cin(a) cin >> a 31 #define FRead() freopen("in", "r", stdin) 32 #define FWrite() freopen("out", "w", stdout) 33 #define Rep(i, len) for(int i = 0; i < (len); i++) 34 #define For(i, a, len) for(int i = (a); i < (len); i++) 35 #define Cls(a) memset((a), 0, sizeof(a)) 36 #define Clr(a, p) memset((a), (p), sizeof(a)) 37 #define Full(a) memset((a), 0x7f7f7f, sizeof(a)) 38 #define lrt rt << 1 39 #define rrt rt << 1 | 1 40 #define pi 3.14159265359 41 #define RT return 42 #define lowbit(p) p & (-p) 43 #define onenum(p) __builtin_popcount(p) 44 typedef long long LL; 45 typedef long double LD; 46 typedef unsigned long long ULL; 47 typedef pair<int, int> pii; 48 typedef pair<string, int> psi; 49 typedef pair<LL, LL> pll; 50 typedef map<string, int> msi; 51 typedef vector<int> vi; 52 typedef vector<LL> vl; 53 typedef vector<vl> vvl; 54 typedef vector<bool> vb; 55 56 const int maxn = 50500; 57 typedef struct Node { 58 int i, n; 59 }Node; 60 61 int n; 62 int a[maxn]; 63 Node b[maxn]; 64 65 bool cmp(Node x, Node y) { 66 if(x.n == y.n) return x.i < y.i; 67 return x.n < y.n; 68 } 69 70 int main() { 71 // FRead(); 72 while(~Rint(n)) { 73 Rep(i, n) { 74 Rint(a[i]); 75 b[i].n = a[i]; b[i].i = i; 76 } 77 sort(b, b+n, cmp); 78 int ret = 0, cur = b[0].i; 79 For(i, 1, n) { 80 if(b[i].i > cur) ret = max(ret, b[i].i-cur); 81 else cur = b[i].i; 82 } 83 printf("%d ", ret); 84 } 85 RT 0; 86 }