【Codevs 1376】帕秋莉•诺蕾姬

http://codevs.cn/problem/1376/

枚举修改哪两位,将sum减去之前位置的数+交换之后  %m==0即可

预处理26的次方+O(n^2)

// <1376.cpp> - Tue Oct 18 21:50:03 2016

// This file is made by YJinpeng，created by XuYike's black technology automatically.

// Copyright (C) 2016 ChangJun High School, Inc.

// I don't know what this program is.

#include <iostream>

#include <vector>

#include <algorithm>

#include <cstring>

#include <cstdio>

#include <cstdlib>

#include <cmath>

using namespace std;

typedef long long LL;

const int MAXN=;

char s[MAXN];int mul[MAXN];

int main()

{

    freopen("1376.in","r",stdin);

    freopen("1376.out","w",stdout);

    scanf("%s",s);

    int n=strlen(s),sum=,m;

    mul[n-]=;scanf("%d",&m);

    for(int i=n-;i;i--)

        (sum+=mul[i]*(s[i]-'A')%m)%=m,mul[i-]=mul[i]*%m;

    (sum+=mul[]*(s[]-'A')%m)%=m;

    if(sum==){printf("0 0");return ;}

    for(int i=;i<n;i++)

        for(int j=i+;j<n;j++){

            if(((sum+(s[i]-'A')*(mul[j]-mul[i]+m)%m)%m+(s[j]-'A')*(mul[i]-mul[j]+m)%m)%m==){

                printf("%d %d",i+,j+);return ;

            }

        }printf("-1 -1");

    return ;

}