标签：cal call 输出 line oar alc 成功 inpu case
Now you have a string consists of uppercase letters, two integers AA and BB. We call a substring wonderful substring when the times it appears in that string is between AA and BB (A \le times \le BA≤times≤B). Can you calculate the number of wonderful substrings in that string?
Input has multiple test cases.
For each line, there is a string SS, two integers AA and BB.
\sum length(S) \le 2 \times 10^6∑length(S)≤2×106,
1 \le A \le B \le length(S)1≤A≤B≤length(S)
For each test case, print the number of the wonderful substrings in a line.
AAA 2 3 ABAB 2 2
2 3
题解：SAM模板题
参考代码：
//H 求子串出现次数在k1=<num<=k2;
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 4e5+10;
char ss[200005];
const int LetterSize = 26;
int tot, last,ch[MAXN][LetterSize],fa[MAXN],len[MAXN];
int sum[MAXN],tp[MAXN],cnt[MAXN];
void init()
{
last = tot = 1;
len[1] = 0;
memset(ch,0,sizeof ch);
memset(fa,0,sizeof fa);
memset(cnt,0,sizeof cnt);
}
void add( int x)
{
int p = last, np = last = ++tot;
len[np] = len[p] + 1, cnt[last] = 1;
while( p && !ch[p][x]) ch[p][x] = np, p = fa[p];
if(p == 0) fa[np] = 1;
else
{
int q = ch[p][x];
if( len[q] == len[p] + 1)
fa[np] = q;
else
{
int nq = ++tot;
memcpy( ch[nq], ch[q], sizeof ch[q]);
len[nq] = len[p] + 1, fa[nq] = fa[q], fa[q] = fa[np] = nq;
while( p && ch[p][x] == q) ch[p][x] = nq, p = fa[p];
}
}
}
void toposort()
{
for(int i = 1; i <= len[last]; i++) sum[i] = 0;
for(int i = 1; i <= tot; i++) sum[len[i]]++;
for(int i = 1; i <= len[last]; i++) sum[i] += sum[i-1];
for(int i = 1; i <= tot; i++) tp[sum[len[i]]--] = i;
}
int main()
{
int k1,k2;
while(scanf("%s",ss)!=EOF)
{
init();
scanf("%d%d",&k1,&k2);
long long ans=0;
for(int i=0,len=strlen(ss);i<len;i++) add(ss[i]-‘A‘);
toposort();
for(int i=tot;i;i--)
{
int p=tp[i],fp=fa[p];
cnt[fp]+=cnt[p];
if(cnt[p]>=k1 && cnt[p]<=k2) ans+=len[p]-len[fp];
}
printf("%lld\n",ans);
}
return 0;
}
ACM-ICPC 2018 焦作赛区网络预赛 H题 String and Times（SAM）
标签：cal call 输出 line oar alc 成功 inpu case
原文地址：https://www.cnblogs.com/songorz/p/9651909.html
