10730-Antiarithmetic?【暴力枚举】

水题

求一个序列是否存在3个数按顺序构成等差数列

直接枚举等差数列的差值 时间复杂度降到 n * n / 3

开pos数组记录每个值得为之

楷vis数组记录目前i是否出现过

强行AC

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 11111;
int array[maxn];
int vis[maxn];
int pos[maxn];
int main(){
    int n;
    while(scanf("%d",&n) && n){
        scanf("%*s");
        memset(vis,0,sizeof(vis));
        for(int i = 0; i < n; i++){
            scanf("%d",&array[i]);
            pos[array[i]] = i;
        }
        int d = n / 3;
        int ok = 0;
        for(int i = 0; i < n; i++){
            vis[array[i]] = 1;
            for(int j = 1; j <= d; j++){
                int e1 = array[i] - 2 * j;
                int e2 = array[i] - j;
                int e3 = array[i] + 2 * j;
                int e4 = array[i] + j;
                if(e1 >= 0){
                    if(vis[e1] && vis[e2] && pos[e1] < pos[e2]){
                        //printf("%d %d %d\n",e1,e2,array[i]);
                        ok = 1;
                        break;
                    }
                }
                if(e2 < n){
                    if(vis[e3] && vis[e4] && pos[e3] < pos[e4]){
                        //printf("%d %d %d\n",e3,e4,array[i]);
                        ok = 1;
                        break;
                    }
                }
            }
            if(ok){
                //printf("%d\n",i);
                break;
            }
        }
        if(ok) printf("no\n");
        else printf("yes\n");
    }
    return 0;
}

10730-Antiarithmetic?【暴力枚举】

原文：http://blog.csdn.net/u013451221/article/details/44655657