bzoj1692

题解:

二分最近的不相同

然后hash判断是否相同

然后贪心

代码:

#include<bits/stdc++.h>
using namespace std;
#define ull unsigned long long 
const int N=31000;
int n;
char s[N],s1[11],ans[N];
ull h1[N],h2[N],pw[N];
int main()
{
    scanf("%d",&n);
    for (int i=1;i<=n;i++)scanf("%s",s1+1),s[i]=s1[1];
    pw[0]=1;
    for (int i=1;i<=n;i++)pw[i]=pw[i-1]*133;
    for (int i=1;i<=n;i++)h1[i]=h1[i-1]*133+s[i];
    for (int i=n;i>=1;i--)h2[i]=h2[i+1]*133+s[i];
    for (int i=1,l1=1,r1=n;i<=n;i++)
     {
        int l=1,r=r1-l1+1;
        while (l<=r)
         {
            int mid=(l+r)>>1;
            if (h1[l1+mid-1]-h1[l1-1]*pw[mid]==h2[r1-mid+1]-h2[r1+1]*pw[mid])l=mid+1;
            else r=mid-1;
         }
        if (l>r1-l1+1)l=1;
        if (s[l1+l-1]>s[r1-l+1])ans[i]=s[r1],r1--;
        else ans[i]=s[l1],l1++;
     }
    for (int i=1;i<=n;i++)
     {
        putchar(ans[i]);
        if (i%80==0)puts("");
     }
    return 0;
}   
原文地址:https://www.cnblogs.com/xuanyiming/p/8510231.html