【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=2010;
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=0,m;
    mul[n-1]=1;scanf("%d",&m);
    for(int i=n-1;i;i--)
        (sum+=mul[i]*(s[i]-'A')%m)%=m,mul[i-1]=mul[i]*26%m;
    (sum+=mul[0]*(s[0]-'A')%m)%=m;
    if(sum==0){printf("0 0");return 0;}
    for(int i=0;i<n;i++)
        for(int j=i+1;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==0){
                printf("%d %d",i+1,j+1);return 0;
            }
        }printf("-1 -1"); 
    return 0;
}