| ||||||||||
| Online Judge | Problem Set | Authors | Online Contests | User | ||||||
|---|---|---|---|---|---|---|---|---|---|---|
| Web Board Home Page F.A.Qs Statistical Charts | Current Contest Past Contests Scheduled Contests Award Contest | |||||||||
你的哈希函数不错啊,借鉴了一下~ 帮你稍微改了下,效率稍稍有所提高In Reply To:TLE的请注意下. Posted by:2008550914 at 2010-04-10 10:35:45 //把求sum部分的循环弄掉了,只要减去开头的加上末尾的就可以减少很多次乘法了
#include<iostream>
using namespace std;
bool hash[16000000]={0};
char text[1000000];
int c[128];
int main(){
int n,nc,i,j=0,length,ans=1,sum=0,tp=1;
scanf("%d%d%s",&n,&nc,text);
length=strlen(text);
for(i=0;text[i];i++){
if(!c[text[i]]) c[text[i]]=++j;
if(j==nc) break;
}
for(i=0;i<n;i++){//预处理一个sum
sum=sum*nc+c[text[i]]-1;
tp*=nc;
}
tp/=nc; hash[sum]=1;
for(i=1;i<=length-n;i++){
sum=(sum-(c[text[i-1]]-1)*tp)*nc+c[text[i+n-1]]-1; //用这个公式替换原来的循环
if(!hash[sum]){
hash[sum]=1;
ans++;
}
}
printf("%d\n",ans);
return 0;
}
Followed by:
Post your reply here: |
All Rights Reserved 2003-2013 Ying Fuchen,Xu Pengcheng,Xie Di
Any problem, Please Contact Administrator