标签:des style blog http io color os ar java
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1075 Accepted Submission(s): 388
题意: chinese
思路:dp[i] 表示以 i 结尾的,连续的最长非递减,
对于 i 位置 ,二分回文的长度,用字符串哈希判断是否可以,把二分的上界设为dp[i] ,就符合非递减要求了,
注意回文串长度有奇数偶数就好了,
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<queue> #include<vector> #include<set> #include<stack> #include<map> #include<ctime> #include<bitset> #define LL long long #define u64 unsigned long long #define yy 137 #define INF 999999 #define maxn 100010 using namespace std; int dp[maxn] ,a[maxn] ; u64 sum1[maxn],sum2[maxn],pow1[maxn]; int main() { int i,j,k,ans; int n,m,T,L,R,mid; u64 tt; pow1[0]=1; for( i = 1 ; i < maxn;i++) pow1[i]=pow1[i-1]*yy; cin >> T ; while(scanf("%d",&n)!= EOF) { for( i = 1 ; i <= n ;i++ ) { scanf("%d",&a[i]) ; sum1[i]=sum1[i-1]+a[i]*pow1[i] ; } k = 1; sum2[n+1]=0; for( i = n ; i >= 1 ;i-- ) { sum2[i]=sum2[i+1]+a[i]*pow1[k++] ; } ans=1; for( i = 1 ; i <= n ;i++) { if(a[i]>=a[i-1])dp[i]=dp[i-1]+1; else dp[i]=1; L=1;R=dp[i]; while(L<=R) { mid=(L+R)>>1; j=n-(mid+i)+1; if(i-mid>=j){ tt=pow1[i-mid-j] ; if(sum1[i]-sum1[i-mid]==tt*(sum2[i]-sum2[i+mid])) { L = mid+1; ans=max(mid*2-1,ans) ; } else R =mid-1; } else{ tt=pow1[j-i+mid] ; if((sum1[i]-sum1[i-mid])*tt==sum2[i]-sum2[i+mid]) { L = mid+1; ans=max(mid*2-1,ans) ; } else R =mid-1; } } L=1;R=dp[i]; // cout << L << " " << while(L<=R) { mid=(L+R)>>1; j=n-(mid+i+1)+1;///// if(i-mid>=j){ tt=pow1[i-mid-j] ; if(sum1[i]-sum1[i-mid]==tt*(sum2[i+1]-sum2[i+1+mid])) { L = mid+1; ans=max(mid*2,ans) ; } else R =mid-1; } else{ tt=pow1[j-i+mid] ; if((sum1[i]-sum1[i-mid])*tt==sum2[i+1]-sum2[i+1+mid]) { L = mid+1; ans=max(mid*2,ans) ; } else R =mid-1; } } } cout << ans<<endl; } return 0; }
标签:des style blog http io color os ar java
原文地址:http://www.cnblogs.com/20120125llcai/p/4058277.html