Codeforces Beta Round #17 C. Balance （字符串计数 dp）
3 seconds
128 megabytes
standard input
standard output
Nick likes strings very much, he likes to rotate them, sort them, rearrange characters within a string... Once he wrote a random string of characters a, b, c on
a piece of paper and began to perform the following operations:
- to take two adjacent characters and replace the second character with the first one,
- to take two adjacent characters and replace the first character with the second one
To understand these actions better, let's take a look at a string «abc». All of the following strings can be obtained by performing one of the described operations
on «abc»: «bbc», «abb»,
«acc». Let's denote the frequency of a character for each of the characters a, band c as
the number of occurrences of this character in the string. For example, for string «abc»: |a|
= 1, |b| = 1, |c| = 1, and for string «bbc»:
|a| = 0, |b| = 2, |c|
= 1.
While performing the described operations, Nick sometimes got balanced strings. Let's say that a string is balanced, if the frequencies of each character differ by at most 1. That is - 1 ≤ |a| - |b| ≤ 1, - 1 ≤ |a| - |c| ≤ 1 и - 1 ≤ |b| - |c| ≤ 1.
Would you help Nick find the number of different balanced strings that can be obtained by performing the operations described above, perhaps multiple times, on the given string s.
This number should be calculated modulo 51123987.
The first line contains integer n (1 ≤ n ≤ 150) —
the length of the given string s. Next line contains the given string s.
The initial string can be balanced as well, in this case it should be counted too. The given string s consists only of characters a, b and c.
Output the only number — the number of different balanced strings that can be obtained by performing the described operations, perhaps multiple times, on the given string s,
modulo 51123987.
4
abca
7
4
abbc
3
2
ab
1
In the first sample it is possible to get 51 different strings through the described operations, but only 7 of
them are balanced: «abca», «bbca», «bcca»,
«bcaa», «abcc», «abbc»,
«aabc». In the second sample: «abbc», «aabc»,
«abcc». In the third sample there is only one balanced string — «ab»
itself.
题意：
给出一个最多150字符长的仅仅有a或b或c组成的字符串
对于每一个操作能够把前面一个字符变成后面一个字符或者把后面的一个字符变成前面一个字符
即能够运行赋值语句 str[i+1] = str[i]; 或者 str[i] = str[i+1];
假设原字符串在运行若干次操作后变成一个a,b,c的字符数量相互不超过1的 字符串, 那么称得到的串为一个合法串
问合法串有多少个。
思路来源于：点击打开链接（推荐，写的好清晰）
思路：
依据性质。不可能交叉替换，如果输入的字符串是 A, 字符串 A'是 A串的同样字符压缩成一个后的结果 ，所以B是由A'一个字符扩展一个区间来的，状态 dp[i][na][nb][nc] 表示 以 A'中第i个字符为结尾的字符a数量为na, 字符b数量为nb, 字符c数量为nc的字符串的数量，防止一个B计算两次，规定下一个字符转移仅仅能用近期的来转移。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define maxn 500005
#define MAXN 100005
#define mod 51123987
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-12
typedef long long ll;
using namespace std; int n,m,ans,cnt,tot,flag;
int dp[155][55][55][55];
int next[155][3];
char s[200],s1[200]; int main()
{
int i,j,k,t;
while(~scanf("%d",&n))
{
scanf("%s",s+1);
tot=1;
s1[1]=s[1];
for(i=2; i<=n; i++)
{
if(s[i]!=s1[tot]) s1[++tot]=s[i];
}
memset(next,0,sizeof(next));
for(j=1; j<=tot; j++)
{
if(s1[j]=='a')
{
if(next[0][0]==0) next[0][0]=j;
}
else if(s1[j]=='b')
{
if(next[0][1]==0) next[0][1]=j;
}
else
{
if(next[0][2]==0) next[0][2]=j;
}
}
for(i=1; i<=tot; i++)
{
for(j=i; j<=tot; j++)
{
if(s1[j]=='a')
{
if(next[i][0]==0) next[i][0]=j;
}
else if(s1[j]=='b')
{
if(next[i][1]==0) next[i][1]=j;
}
else
{
if(next[i][2]==0) next[i][2]=j;
}
}
}
int na,nb,nc;
m=n/3+1;
dp[0][0][0][0]=1;
for(i=0; i<=tot; i++)
{
for(na=0; na<=m; na++)
{
for(nb=0; nb<=m; nb++)
{
for(nc=0; nc<=m; nc++)
{
if(next[i][0])
{
dp[next[i][0]][na+1][nb][nc]+=dp[i][na][nb][nc];
dp[next[i][0]][na+1][nb][nc]%=mod;
}
if(next[i][1])
{
dp[next[i][1]][na][nb+1][nc]+=dp[i][na][nb][nc];
dp[next[i][1]][na][nb+1][nc]%=mod;
}
if(next[i][2])
{
dp[next[i][2]][na][nb][nc+1]+=dp[i][na][nb][nc];
dp[next[i][2]][na][nb][nc+1]%=mod;
}
}
}
}
}
ans=0;
if(n%3==0)
{
for(i=1; i<=tot; i++)
{
ans+=dp[i][n/3][n/3][n/3];
ans%=mod;
}
}
else if(n%3==1)
{
for(i=1; i<=tot; i++)
{
ans+=dp[i][n/3+1][n/3][n/3];
ans+=dp[i][n/3][n/3+1][n/3];
ans+=dp[i][n/3][n/3][n/3+1];
ans%=mod;
}
}
else
{
for(i=1; i<=tot; i++)
{
ans+=dp[i][n/3+1][n/3+1][n/3];
ans+=dp[i][n/3+1][n/3][n/3+1];
ans+=dp[i][n/3][n/3+1][n/3+1];
ans%=mod;
}
}
printf("%d\n",ans);
}
return 0;
}
/*
4
abca
*/
最新文章
- 【译】Spring 4 基于TaskScheduler实现定时任务（注解）
- 原生AJAX入门讲解（含实例）
- Codeforces Round #377 (Div. 2) B. Cormen — The Best Friend Of a Man（贪心）
- table常用功能总结
- winform中listview imagelist问题
- Swiper.js 中文API手册
- 用 unoconv 将 xls 转换成 csv
- Clang Language Extensions
- Javascript禁止网页复制粘贴效果，或者复制时自动添加来源信息
- jsoup技术抓取网页数据大全
- 仿猪八戒一个提示(jQuery插件) v0.1 beta
- 转csdn-css4
- centos7 网卡识别为eth0 eth1
- step_by_step_Angularjs-UI-Grid使用简介
- 存储过程中的 SET XACT_ABORT ON 和事务
- ubuntu后台运行命令
- 面试题----gcc的编译流程
- 用sparkR， 分析上亿条订单数据的脚本。
- 五、搭建kube-dns
- 2017-2018-1 20155330 《信息安全系统设计基础》加分项目--实现mypwd