Codeforces 39J Spelling Check hash

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题目链接：点击打开链接

题意：

给定2个字符串

选择第一个字符串的其中一个字母删除，使得2个字符串完全相同

问哪些位置可以选

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;
}

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Codeforces 39J Spelling Check hash

标签：blog   http   os   2014   io   for   

原文地址：http://blog.csdn.net/qq574857122/article/details/37881757