主题链接:点击打开链接
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?'
':' ');
}
return 0;
}
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