Balance CodeForces - 17C

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最新推荐文章于 2019-11-09 16:23:52 发布

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分类专栏： # 基础DP(动态规划)

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本文介绍了一种算法，用于计算通过特定操作从给定字符串出发能得到的不同平衡字符串的数量。平衡字符串是指其中字符a、b、c的数量相差不超过1的字符串。

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Balance

CodeForces - 17C

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, b and 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.

Input

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

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.

Example

Input

4
abca

Output

Input

4
abbc

Output

Input

2
ab

Output

Note

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.

代码转载至： http://blog.youkuaiyun.com/Yunyouxi/article/details/24009567

题目意思

给出一个最多150字符长的只有a或b或c组成的字符串

对于每个操作可以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

即可以执行赋值语句 str[i+1] = str[i]; 或者 str[i] = str[i+1];

如果原字符串在执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

例如输入字符串 aaacb 其中a有3个,b有1个,c有1个 3-1>1 不合法

但是可以把第3个a变成 c -> aaccb a有2个,b有1个,c有2个, 相互之间的数量都不超过1 所以 aaccb 就是一个合法的串

解题方法

假设输入的字符串是 A, 字符串 A'是 A串的相同字符压缩成一个后的结果 (例如 aacbbb 压缩成 acb)

假设一个合法串B, B'是B压缩后的字符串

这时会发现如果 B 是由 A 经过若干次操作得到的结果, 那么 B' 是 A'的一个子序列

如果 B' 是 A' 的一个子序列, 那么说明 B 能从 A 通过若干次操作得到

这样很自然可以想到如果用一个数组 dp[i] 表示以 A'中的第 i 个字符为结尾的串的数量

由于题目要求字符a, b, c的数量相互不超过1 所以要同时记录字符a, b, c使用了的数量即

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

由于A'中只有排在前面的字符才能更新后面的字符所以不断用当前状态去更新后面的状态当枚举到后面的状态时该状态已经是最优的了

更新时用完全背包的思想, 即可以在当前状态后面加任意个的a 或 b 或 c (详情看代码)

if(构造的下一个字符是a) dp[next[i]][na+1][nb][nc] += dp[i][na][nb][nc]; next[i]表示满足条件(A'[j] == 'a' && j >= i)中最小的 j

if(构造的下一个字符是b) dp[next[i]][na][nb+1][nc] += dp[i][na][nb][nc]; next[i]表示满足条件(A'[j] == 'b' && j >= i)中最小的 j

if(构造的下一个字符是c) dp[next[i]][na][nb][nc+1] += dp[i][na][nb][nc]; next[i]表示满足条件(A'[j] == 'c' && j >= i)中最小的 j

为什么是最小的 j ? 如果不是最小的 j 是不是会使一部分字符浪费了?

由于要求合法串的a,b,c字符数量相互不超过1 如果字符串长度 n % 3 == 0 则a,b,c字符数量均为 n/3

如果字符串长度 n % 3 == 1 则a,b,c字符数量可能为 n/3+1,n/3,n/3 或 n/3,n/3+1,n/3 或 n/3,n/3,n/3+1

如果字符串长度 n % 3 == 2 则a,b,c字符数量可能为 n/3,n/3+1,n/3+1 或 n/3+1,n/3,n/3+1 或 n/3+1,n/3+1,n/3

即 na, nb, nc的最大值只是50

所以时间复杂度大约为 150*50*50*50 = 18750000 不会超时

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;
const int MAXN = 1e3 + 10;
const int MOD = 51123987;
char str[MAXN];
char s[MAXN];
int dp[200][53][53][53];
int nex[200][3];
int main(){
    int n;
    while(~scanf("%d",&n)){
        memset(dp,0,sizeof(dp));
        scanf("%s",str);
        int k = 0;
        s[k++] = str[0];
        for(int i = 1; i < n; i++){
            if(str[i] != str[i-1])
                s[k++] = str[i];
        }
        memset(nex,-1,sizeof(nex));
        for(int i = 0; i < k; i++){
            for(int j = i; j < k; j++){
                if(s[j] == 'a' && nex[i][0] == -1) nex[i][0] = j;
                if(s[j] == 'b' && nex[i][1] == -1) nex[i][1] = j;
                if(s[j] == 'c' && nex[i][2] == -1) nex[i][2] = j;
            }
        }
        int tn = n / 3 + 1;
        dp[0][0][0][0] = 1;
        for(int i = 1; i <= k; i++){
            for(int A = 0; A <= tn; A++){
                for(int B = 0; B <= tn; B++){
                    for(int C = 0; C <= tn; C++){
                        if(dp[i-1][A][B][C] == 0) continue;
                        int dex = nex[i-1][0];
                        if(dex != -1){
                            dp[dex][A+1][B][C] += dp[i-1][A][B][C];
                            dp[dex][A+1][B][C] %= MOD;
                        }
                        dex = nex[i-1][1];
                        if(dex != -1){
                            dp[dex][A][B+1][C] += dp[i-1][A][B][C];
                            dp[dex][A][B+1][C] %= MOD;
                        }
                        dex = nex[i-1][2];
                        if(dex != -1){
                            dp[dex][A][B][C+1] += dp[i-1][A][B][C];
                            dp[dex][A][B][C+1] %= MOD;
                        }
                    }
                }
            }
        }
        int res = 0;
        if(n % 3 == 0){
            for(int i = 0; i < k; i++){
                res += dp[i][n/3][n/3][n/3];
                res %= MOD;
            }
        }
        else if(n % 3 == 1){
            for(int i = 0; i < k; i++){
                res += dp[i][n/3+1][n/3][n/3];
                res %= MOD;
                res += dp[i][n/3][n/3+1][n/3];
                res %= MOD;
                res += dp[i][n/3][n/3][n/3+1];
                res %= MOD;
            }
        }
        else{
            for(int i = 0; i < k; i++){
                res += dp[i][n/3+1][n/3+1][n/3];
                res %= MOD;
                res += dp[i][n/3+1][n/3][n/3+1];
                res %= MOD;
                res += dp[i][n/3][n/3+1][n/3+1];
                res %= MOD;
            }
        }
        printf("%d\n",res);
    }
    return 0;
}