No key code will be wa
abcdefg
1
adcabcabcdefg
Well understood abc next traversed a non-compliance cnt = 0;
while traversing from next next start will start from abcddefg in b
and want to start from a need to reduce if j- j, then sank into the cycle of death
at this time We choose j-cnt, although it will traverse from the second letter of the string that has just been traversed (one more traversal), but it has a very good effect. Currently, there is no better way to solve this problem.
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
char a[111111];
int n;
char s[111111];
int ans;
int cnt;
int v[111111], t;//用v数组存所有的cnt结果
int maxs;//用 maxs更新最大的cnt
bool cmp(int a, int b)
{
return a > b;
}
int main()
{
scanf("%s", a);
cin >> n;
for(int i = 0; i < n; i++)
{
scanf("%s",s);
cnt = 0;
maxs = 0;
memset(v, 0, sizeof(v));
t = 0;
for(int j = 0; j < strlen(s); j++)
{
cout << "j = " << j << "\n";
if(a[cnt] == s[j]){
cnt++;
//cout << s[i] << "\n";
v[t++] = cnt;
maxs = max(maxs, cnt);
}
else{
v[t++] = cnt;
j = j - cnt; // 关键 一直wa
//cout << "j = " << j << "\n";
cnt = 0;
}
}
sort(v, v + t ,cmp);
//printf("v = %d\n",v[0]);
ans += v[0];// 或者 把v[0]改成maxs
}
cout << ans << "\n";
return 0;
}