这两题都是统计一个字符串中另一个字符串的数量,只不过一题可以重叠,另一题不能重叠。
在代码上出了输入格式,也就是一个语句的区别,其实就是找到了一个文本串和模式串匹配的子串之后,是不是从模式串开头重新找的区别。
第一题AC代码:
/*
* @Author: hesorchen
* @Date: 2020-07-02 22:19:34
* @LastEditTime: 2020-07-11 12:49:56
* @Description: https://hesorchen.github.io/
*/
#include <map>
#include <set>
#include <list>
#include <queue>
#include <deque>
#include <cmath>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define endl '\n'
#define PI acos(-1)
#define PB push_back
#define ll long long
#define INF 0x3f3f3f3f
#define mod 1000000007
#define pll pair<ll, ll>
#define lowbit(abcd) (abcd & (-abcd))
#define max(a, b) ((a > b) ? (a) : (b))
#define min(a, b) ((a < b) ? (a) : (b))
#define IOS \
ios::sync_with_stdio(false); \
cin.tie(0); \
cout.tie(0);
#define FRE \
{ \
freopen("in.txt", "r", stdin); \
freopen("out.txt", "w", stdout); \
}
inline ll read()
{
ll x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch ^ 48);
ch = getchar();
}
return x * f;
}
//==============================================================================
ll nxt[1000010];
char a[1000010], b[1000010];
void get_nxt()
{
ll lenb = strlen(b);
int j = 0, k = -1;
nxt[0] = -1;
while (j < lenb)
{
if (k == -1 || b[j] == b[k])
nxt[++j] = ++k;
else
k = nxt[k];
}
}
int KMP()
{
ll lenb = strlen(b), lena = strlen(a), res = 0;
get_nxt();
ll i = 0, j = 0;
while (i < lena)
{
if (j == -1 || a[i] == b[j])
i++, j++;
else
j = nxt[j];
if (j == lenb)
{
res++;
// 能不能重复
// j = nxt[j];
j = 0;
}
}
return res;
}
int main()
{
IOS;
ll n;
while (cin >> a)
{
if (a[0] == '#')
break;
cin >> b;
cout << KMP() << endl;
}
return 0;
}
/*
ababa
aba
*/
第二题AC代码:
/*
* @Author: hesorchen
* @Date: 2020-07-02 22:19:34
* @LastEditTime: 2020-07-11 12:58:25
* @Description: https://hesorchen.github.io/
*/
#include <map>
#include <set>
#include <list>
#include <queue>
#include <deque>
#include <cmath>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define endl '\n'
#define PI acos(-1)
#define PB push_back
#define ll long long
#define INF 0x3f3f3f3f
#define mod 1000000007
#define pll pair<ll, ll>
#define lowbit(abcd) (abcd & (-abcd))
#define max(a, b) ((a > b) ? (a) : (b))
#define min(a, b) ((a < b) ? (a) : (b))
#define IOS \
ios::sync_with_stdio(false); \
cin.tie(0); \
cout.tie(0);
#define FRE \
{ \
freopen("in.txt", "r", stdin); \
freopen("out.txt", "w", stdout); \
}
inline ll read()
{
ll x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
{
x = (x << 1) + (x << 3) + (ch ^ 48);
ch = getchar();
}
return x * f;
}
//==============================================================================
ll nxt[1000010];
char a[1000010], b[1000010];
void get_nxt()
{
ll lenb = strlen(b);
int j = 0, k = -1;
nxt[0] = -1;
while (j < lenb)
{
if (k == -1 || b[j] == b[k])
nxt[++j] = ++k;
else
k = nxt[k];
}
}
int KMP()
{
ll lenb = strlen(b), lena = strlen(a), res = 0;
get_nxt();
ll i = 0, j = 0;
while (i < lena)
{
if (j == -1 || a[i] == b[j])
i++, j++;
else
j = nxt[j];
if (j == lenb)
{
res++;
// 能不能重复
j = nxt[j];
// j = 0;
}
}
return res;
}
int main()
{
IOS;
ll n;
cin >> n;
while (n--)
{
if (a[0] == '#')
break;
cin >> b >> a;
cout << KMP() << endl;
}
return 0;
}
/*
ababa
aba
*/