10730-Antiarithmetic?【暴力枚举】

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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

1522139710730Antiarithmetic?AcceptedC++0.0352015-03-26 12:09:56

#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;}

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