本文共 2010 字，大约阅读时间需要 6 分钟。
和51nod1055 一样；
#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include //#include //#pragma GCC optimize(2)using namespace std;#define maxn 5005#define inf 0x7fffffff//#define INF 1e18#define rdint(x) scanf("%d",&x)#define rdllt(x) scanf("%lld",&x)#define rdult(x) scanf("%lu",&x)#define rdlf(x) scanf("%lf",&x)#define rdstr(x) scanf("%s",x)#define mclr(x,a) memset((x),a,sizeof(x))typedef long long ll;typedef unsigned long long ull;typedef unsigned int U;#define ms(x) memset((x),0,sizeof(x))const long long int mod = 1e9 + 7;#define Mod 1000000000#define sq(x) (x)*(x)#define eps 1e-5typedef pair pii;#define pi acos(-1.0)//const int N = 1005;#define REP(i,n) for(int i=0;i<(n);i++)typedef pair pii;inline int rd() { int x = 0; char c = getchar(); bool f = false; while (!isdigit(c)) { if (c == '-') f = true; c = getchar(); } while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); } return f ? -x : x;}ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a%b);}int sqr(int x) { return x * x; }/*ll ans;ll exgcd(ll a, ll b, ll &x, ll &y) { if (!b) { x = 1; y = 0; return a; } ans = exgcd(b, a%b, x, y); ll t = x; x = y; y = t - a / b * y; return ans;}*/int n;int a[maxn];int dp[5005][5005];int main(){ // ios::sync_with_stdio(0); n = rd(); for (int i = 1; i <= n; i++) { a[i] = rd(); } sort(a + 1, a + 1 + n); for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++)dp[i][j] = 2; } int ans = 2; for (int i = 1; i <= n - 1; i++) { int pre = i - 1; for (int j = i + 1; j <= n; j++) { while (pre > 0 && a[j] - a[i] > a[i] - a[pre])pre--; if (!pre)break; if (pre > 0 && a[j] - a[i] == a[i] - a[pre]) dp[j][i] = max(dp[j][i], dp[i][pre] + 1); ans = max(ans, dp[j][i]); } } cout << ans << endl; return 0;}
转载于:https://www.cnblogs.com/zxyqzy/p/10375494.html
