HDU5328.Problem Killer

来源：互联网发布：淘宝客怎么找免单商品编辑：程序博客网时间：2024/05/22 01:43

题解:
首先如果a,b,c是等差数列,且b,c,d也是等差数列,那么a,b,c,d也是等差数列,这个结论等于等比数列也成立，所以只要对于每个i(i+2

#include <map>#include <set>#include <stack>#include <queue>#include <cmath>#include <ctime>#include <vector>#include <cstdio>#include <cctype>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;#define INF 0x3f3f3f3f#define inf -0x3f3f3f3f#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define mem0(a) memset(a,0,sizeof(a))#define mem1(a) memset(a,-1,sizeof(a))#define mem(a, b) memset(a, b, sizeof(a))typedef long long ll;const int maxn=1e6+10;int a[maxn];int main(){    int t;    int n;    //freopen("1002.in","r",stdin);    //freopen("1001.out","w",stdout);    scanf("%d",&t);    while(t--){        scanf("%d",&n);        for(int i=0;i<n;i++)            scanf("%d",&a[i]);        if(n==1){            printf("1\n");             continue;        }        int ans=1;        for(int i=0;i<n;){            int count1=2;            if((i+2)<n&&a[i]+a[i+2]==2*a[i+1])                while((i+2)<n&&a[i]+a[i+2]==2*a[i+1]){                    i++;                    count1++;                }            else                i++;            ans=max(count1,ans);        }        for(int i=0;i<n;){            int count1=2;            if((i+2)<n&&(ll)a[i]*a[i+2]==(ll)a[i+1]*a[i+1])                while((i+2)<n&&(ll)a[i]*a[i+2]==(ll)a[i+1]*a[i+1]){                    i++;                    count1++;                }            else                i++;            ans=max(count1,ans);        }        printf("%d\n",ans);    }    return 0;}

0 0