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意甲冠军：

特定2弦

选择中删除一个字符串的第一个字母，得2个字符串全然同样

问哪些位置能够选

思路：

hash求前缀后缀。然后枚举位置

#include <cstdio>

#include <algorithm>

#include<iostream>

#include<string.h>

#include <math.h>

#include<queue>

#include<map>

#include<vector>

#include<set>

using namespace std;

#define mod 100000007

#define ll long long

#define N 1000050

char s[N], c[N];

vector <int> ans;

ll l1[N], r1[N], l2[N], r2[N];

int main(){

    int i, j;

    while(~scanf("%s",s+1)){

        scanf("%s",c+1);

        int len1 = strlen(s+1), len2 = strlen(c+1);

        if(len1 -1 != len2){puts("0");continue;}

        l1[0] = 0;

        for(i = 1; i <= len1; i++) {

            l1[i] = l1[i-1]*26 + s[i];

            if(l1[i]>=mod) l1[i] %= mod;

        }

        r1[len1+1] = 0;

        for(i = len1; i ; i--) {

            r1[i] = r1[i+1]*26 + s[i];

            if(r1[i]>=mod) r1[i] %= mod;

        }

        l2[0] = 0;

        for(i = 1; i <= len2; i++) {

            l2[i] = l2[i-1]*26 + c[i];

            if(l2[i]>=mod) l2[i] %= mod;

        }

        r2[len2+1] = 0;

        for(i = len2; i ; i--) {

            r2[i] = r2[i+1]*26 + c[i];

            if(r2[i]>=mod) r2[i] %= mod;

        }

        ans.clear();

        for(i = 1; i <= len1; i++) {

            ll a = l1[i-1] + r1[i+1];

            ll b = l2[i-1]+r2[i];

            if(a==b)ans.push_back(i);

        }

        cout<<ans.size()<<endl;

        for(i = 0; i <ans.size(); i++)

            printf("%d%c", ans[i], i==ans.size()-1?'\n':' ');

    }

    return 0;

}