String

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类似Hdu3336

给出一个字符串，请算出它的每个前缀分别在字符串中出现了多少次。再将这些结果加起来输出
Input
The first line include a number T, means the number of test cases.
For each test case, just a line only include lowercase indicate the String S, the length of S will less than 100000.
Output
For each test case, just a number means the sum.
Sample Input

3
acacm
moreandmorecold
thisisthisththisisthisisthisththisis

Sample Output

7
19
82
HINT
For the first case,
there are two "a","ac" and one "aca","acac","acacm" in the string "acacm".
So the answer is 2 + 2 + 1 + 1 + 1 = 7

Sol1:利用Kmp的性质,暴力统计,但是会TLE

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll o,p[300011],sum[300011],len,ans,j;
char a[300011];
int main()
{
	scanf("%lld",&o);
	while(o--)
	{
		memset(p,0,sizeof(p));
		memset(sum,0,sizeof(sum));
		scanf("%s",a+1);
		len=strlen(a+1);
		
		ans=0;j=0;
		for(ll i=1;i<len;i++)
		{
			while(j&&a[j+1]!=a[i+1])
			    j=p[j];
			if(a[j+1]==a[i+1])
			    j++;
			p[i+1]=j;
		}
		for(ll i=1;i<=len;i++)
		{
		    sum[i]=1;
		    ll jj=i;
		    while (p[jj]!=0)
		    {
		    	sum[i]++;
		    	jj=p[jj];
		    }
		    	
		    ans=ans+sum[i];
	    }
		printf("%lld\n",ans);
	}
	return 0;
}

Sol2:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll o,p[100011],sum[100011],len,ans,j;
char a[100011];
int main()
{
    scanf("%lld",&o);
    while(o--)
    {
        memset(p,0,sizeof(p));
        memset(sum,0,sizeof(sum));
        scanf("%s",a+1);
        len=strlen(a+1);
        ans=0;j=0;
        for(ll i=1;i<len;i++)
        {
            while(j&&a[j+1]!=a[i+1])
                j=p[j];
            if(a[j+1]==a[i+1])
                j++;
            p[i+1]=j;
        }
        for(ll i=1;i<=len;i++)
            sum[i]=sum[p[i]]+1,
            ans+=sum[i];
        printf("%lld\n",ans);
    }
    return 0;
}

String

标签：def   name   nbsp   names   code   sum   输出   his   暴力   

原文地址：https://www.cnblogs.com/cutemush/p/12505574.html